Is the sizeof some pointer always equal to four

Question

For example  sizeof char   returns 4   As does int   long long   everything that I ve tried   Are there any exceptions to this

User · Answer

Just for completeness and historic interest, in the 64bit world there were different platform conventions on the sizes of long and long long types, named LLP64 and LP64, mainly between Unix-type systems and Windows. An old standard named ILP64 also made int = 64-bit wide.

Microsoft maintained LLP64 where longlong = 64 bit wide, but long remained at 32, for easier porting.

Type           ILP64   LP64   LLP64
char              8      8       8
short            16     16      16
int              64     32      32
long             64     64      32
long long        64     64      64
pointer          64     64      64

Source: https://stackoverflow.com/a/384672/48026

User · Answer

In general  sizeof pretty much anything  will change when you compile on different platforms  On a 32 bit platform  pointers are always the same size  On other platforms  64 bit being the obvious example  this can change

User · Answer

In general  sizeof pretty much anything  will change when you compile on different platforms  On a 32 bit platform  pointers are always the same size  On other platforms  64 bit being the obvious example  this can change

User · Answer

The size of the pointer basically depends on the architecture of the system in which it is implemented  For example the size of a pointer in 32 bit is 4 bytes  32 bit   and 8 bytes 64 bit    in a 64 bit machines  The bit types in a machine are nothing but memory address  that it can have  32 bit machines can have 2 32 address space and 64 bit machines can have upto 2 64 address spaces  So a pointer  variable which points to a memory location  should be able to point to any of the memory address  2 32 for 32 bit and 2 64 for 64 bit  that a machines holds   Because of this reason we see the size of a pointer to be 4 bytes in 32 bit machine and 8 bytes in a 64 bit machine

User · Answer

From what I recall  it s based on the size of a memory address  So on a system with a 32-bit address scheme  sizeof will return 4  since that s 4 bytes

User · Answer

if you are compiling for a 64-bit machine  then it may be 8

User · Answer

In Win64  Cygwin GCC 5 4   let s see the below example   First  test the following struct   struct list node      int a      list node  prev      list node  next      struct test struc      char a  b       The test code is below   std  cout lt  lt  sizeof int                lt  lt sizeof int  lt  lt std  endl  std  cout lt  lt  sizeof int                lt  lt sizeof int   lt  lt std  endl  std  cout lt  lt std  endl   std  cout lt  lt  sizeof double             lt  lt sizeof double  lt  lt std  endl  std  cout lt  lt  sizeof double             lt  lt sizeof double   lt  lt std  endl  std  cout lt  lt std  endl   std  cout lt  lt  sizeof list node          lt  lt sizeof list node  lt  lt std  endl  std  cout lt  lt  sizeof list node          lt  lt sizeof list node   lt  lt std  endl  std  cout lt  lt std  endl   std  cout lt  lt  sizeof test struc         lt  lt sizeof test struc  lt  lt std  endl  std  cout lt  lt  sizeof test struc         lt  lt sizeof test struc   lt  lt std  endl        The output is below   sizeof int              4 sizeof int              8  sizeof double           8 sizeof double           8  sizeof list node        24 sizeof list node        8  sizeof test struc       2 sizeof test struc       8   You can see that in 64-bit  sizeof pointer  is 8

User · Answer

Just another exception to the already posted list  On 32-bit platforms  pointers can take 6  not 4  bytes    include  lt stdio h gt   include  lt stdlib h gt   int main         char far  ptr     note that this is a far pointer     printf    d n   sizeof  ptr        return EXIT SUCCESS      If you compile this program with Open Watcom and run it  you ll get 6  because far pointers that it supports consist of 32-bit offset and 16-bit segment values

User · Answer

if you are compiling for a 64-bit machine  then it may be 8

User · Answer

A pointer is just a container for an address   On a 32 bit machine  your address range is 32 bits  so a pointer will always be 4 bytes   On a 64 bit machine were you have an address range of 64 bits  a pointer will be 8 bytes

User · Answer

No  the size of a pointer may vary depending on the architecture  There are numerous exceptions

User · Answer

Even on a plain x86 32 bit platform  you can get a variety of pointer sizes  try this out for an example   struct A      struct B   virtual public A      struct C      struct D   public A  public C      int main         cout  lt  lt   A    lt  lt  sizeof void  A         lt  lt  endl      cout  lt  lt   B    lt  lt  sizeof void  B         lt  lt  endl      cout  lt  lt   D    lt  lt  sizeof void  D         lt  lt  endl      Under Visual C   2008  I get 4  12 and 8 for the sizes of the pointers-to-member-function   Raymond Chen talked about this here

User · Answer

The guarantee you get is that sizeof char     1   There are no other guarantees  including no guarantee that sizeof int       sizeof double      In practice  pointers will be size 2 on a 16-bit system  if you can find one   4 on a 32-bit system  and 8 on a 64-bit system  but there s nothing to be gained in relying on a given size

User · Answer

In addition to what people have said about 64-bit  or whatever  systems  there are other kinds of pointer than pointer-to-object    A pointer-to-member might be almost any size  depending how they re implemented by your compiler  they aren t necessarily even all the same size  Try a pointer-to-member of a POD class  and then a pointer-to-member inherited from one of the base classes of a class with multiple bases  What fun

User · Answer

Just another exception to the already posted list  On 32-bit platforms  pointers can take 6  not 4  bytes    include  lt stdio h gt   include  lt stdlib h gt   int main         char far  ptr     note that this is a far pointer     printf    d n   sizeof  ptr        return EXIT SUCCESS      If you compile this program with Open Watcom and run it  you ll get 6  because far pointers that it supports consist of 32-bit offset and 16-bit segment values

User · Answer

In addition to the 16 32 64 bit differences even odder things can occur   There have been machines where sizeof int    will be one value  probably 4 but where sizeof char    is larger   Machines that naturally address words instead of bytes have to  augment  character pointers to specify what portion of the word you really want in order to properly implement the C C   standard   This is now very unusual as hardware designers have learned the value of byte addressability

User · Answer

In Win64  Cygwin GCC 5 4   let s see the below example   First  test the following struct   struct list node      int a      list node  prev      list node  next      struct test struc      char a  b       The test code is below   std  cout lt  lt  sizeof int                lt  lt sizeof int  lt  lt std  endl  std  cout lt  lt  sizeof int                lt  lt sizeof int   lt  lt std  endl  std  cout lt  lt std  endl   std  cout lt  lt  sizeof double             lt  lt sizeof double  lt  lt std  endl  std  cout lt  lt  sizeof double             lt  lt sizeof double   lt  lt std  endl  std  cout lt  lt std  endl   std  cout lt  lt  sizeof list node          lt  lt sizeof list node  lt  lt std  endl  std  cout lt  lt  sizeof list node          lt  lt sizeof list node   lt  lt std  endl  std  cout lt  lt std  endl   std  cout lt  lt  sizeof test struc         lt  lt sizeof test struc  lt  lt std  endl  std  cout lt  lt  sizeof test struc         lt  lt sizeof test struc   lt  lt std  endl        The output is below   sizeof int              4 sizeof int              8  sizeof double           8 sizeof double           8  sizeof list node        24 sizeof list node        8  sizeof test struc       2 sizeof test struc       8   You can see that in 64-bit  sizeof pointer  is 8

User · Answer

A pointer is just a container for an address   On a 32 bit machine  your address range is 32 bits  so a pointer will always be 4 bytes   On a 64 bit machine were you have an address range of 64 bits  a pointer will be 8 bytes

User · Answer

A pointer is just a container for an address   On a 32 bit machine  your address range is 32 bits  so a pointer will always be 4 bytes   On a 64 bit machine were you have an address range of 64 bits  a pointer will be 8 bytes

User · Answer

From what I recall  it s based on the size of a memory address  So on a system with a 32-bit address scheme  sizeof will return 4  since that s 4 bytes

User · Answer

In general  sizeof pretty much anything  will change when you compile on different platforms  On a 32 bit platform  pointers are always the same size  On other platforms  64 bit being the obvious example  this can change

User · Answer

A pointer is just a container for an address   On a 32 bit machine  your address range is 32 bits  so a pointer will always be 4 bytes   On a 64 bit machine were you have an address range of 64 bits  a pointer will be 8 bytes

User · Answer

8 bit and 16 bit pointers are used in most low profile microcontrollers  That means every washing machine  micro  fridge  older TVs  and even cars   You could say these have nothing to do with real world programming  But here is one real world example  Arduino with 1-2-4k ram  depending on chip  with 2 byte pointers   It s recent  cheap  accessible for everyone and worths coding for

User · Answer

Even on a plain x86 32 bit platform  you can get a variety of pointer sizes  try this out for an example   struct A      struct B   virtual public A      struct C      struct D   public A  public C      int main         cout  lt  lt   A    lt  lt  sizeof void  A         lt  lt  endl      cout  lt  lt   B    lt  lt  sizeof void  B         lt  lt  endl      cout  lt  lt   D    lt  lt  sizeof void  D         lt  lt  endl      Under Visual C   2008  I get 4  12 and 8 for the sizes of the pointers-to-member-function   Raymond Chen talked about this here

User · Answer

From what I recall  it s based on the size of a memory address  So on a system with a 32-bit address scheme  sizeof will return 4  since that s 4 bytes

User · Answer

In general  sizeof pretty much anything  will change when you compile on different platforms  On a 32 bit platform  pointers are always the same size  On other platforms  64 bit being the obvious example  this can change

User · Answer

Just for completeness and historic interest, in the 64bit world there were different platform conventions on the sizes of long and long long types, named LLP64 and LP64, mainly between Unix-type systems and Windows. An old standard named ILP64 also made int = 64-bit wide.

Microsoft maintained LLP64 where longlong = 64 bit wide, but long remained at 32, for easier porting.

Type           ILP64   LP64   LLP64
char              8      8       8
short            16     16      16
int              64     32      32
long             64     64      32
long long        64     64      64
pointer          64     64      64

Source: https://stackoverflow.com/a/384672/48026

User · Answer

No  the size of a pointer may vary depending on the architecture  There are numerous exceptions

User · Answer

The reason the size of your pointer is 4 bytes is because you are compiling for a 32-bit architecture   As FryGuy pointed out  on a 64-bit architecture you would see 8

User · Answer

Size of pointer and int is 2 bytes in Turbo C compiler on windows 32 bit machine    So size of pointer is compiler specific  But generally most of the compilers are implemented to support 4 byte pointer variable in 32 bit and 8 byte pointer variable in 64 bit machine     So size of pointer is not same in all machines

User · Answer

The guarantee you get is that sizeof char     1   There are no other guarantees  including no guarantee that sizeof int       sizeof double      In practice  pointers will be size 2 on a 16-bit system  if you can find one   4 on a 32-bit system  and 8 on a 64-bit system  but there s nothing to be gained in relying on a given size

User · Answer

Technically speaking  the C standard only guarantees that sizeof char     1  and the rest is up to the implementation  But on modern x86 architectures  e g  Intel AMD chips  it s fairly predictable   You ve probably heard processors described as being 16-bit  32-bit  64-bit  etc  This usually means that the processor uses N-bits for integers  Since pointers store memory addresses  and memory addresses are integers  this effectively tells you how many bits are going to be used for pointers  sizeof is usually measured in bytes  so code compiled for 32-bit processors will report the size of pointers to be 4  32 bits   8 bits per byte   and code for 64-bit processors will report the size of pointers to be 8  64 bits   8 bits per byte   This is where the limitation of 4GB of RAM for 32-bit processors comes from -- if each memory address corresponds to a byte  to address more memory you need integers larger than 32-bits

User · Answer

Just another exception to the already posted list  On 32-bit platforms  pointers can take 6  not 4  bytes    include  lt stdio h gt   include  lt stdlib h gt   int main         char far  ptr     note that this is a far pointer     printf    d n   sizeof  ptr        return EXIT SUCCESS      If you compile this program with Open Watcom and run it  you ll get 6  because far pointers that it supports consist of 32-bit offset and 16-bit segment values

User · Answer

if you are compiling for a 64-bit machine  then it may be 8

User · Answer

if you are compiling for a 64-bit machine  then it may be 8

User · Answer

In addition to the 16 32 64 bit differences even odder things can occur   There have been machines where sizeof int    will be one value  probably 4 but where sizeof char    is larger   Machines that naturally address words instead of bytes have to  augment  character pointers to specify what portion of the word you really want in order to properly implement the C C   standard   This is now very unusual as hardware designers have learned the value of byte addressability

User · Answer

The reason the size of your pointer is 4 bytes is because you are compiling for a 32-bit architecture   As FryGuy pointed out  on a 64-bit architecture you would see 8

User · Answer

In addition to what people have said about 64-bit  or whatever  systems  there are other kinds of pointer than pointer-to-object    A pointer-to-member might be almost any size  depending how they re implemented by your compiler  they aren t necessarily even all the same size  Try a pointer-to-member of a POD class  and then a pointer-to-member inherited from one of the base classes of a class with multiple bases  What fun

User · Answer

8 bit and 16 bit pointers are used in most low profile microcontrollers  That means every washing machine  micro  fridge  older TVs  and even cars   You could say these have nothing to do with real world programming  But here is one real world example  Arduino with 1-2-4k ram  depending on chip  with 2 byte pointers   It s recent  cheap  accessible for everyone and worths coding for

User · Answer

The size of the pointer basically depends on the architecture of the system in which it is implemented  For example the size of a pointer in 32 bit is 4 bytes  32 bit   and 8 bytes 64 bit    in a 64 bit machines  The bit types in a machine are nothing but memory address  that it can have  32 bit machines can have 2 32 address space and 64 bit machines can have upto 2 64 address spaces  So a pointer  variable which points to a memory location  should be able to point to any of the memory address  2 32 for 32 bit and 2 64 for 64 bit  that a machines holds   Because of this reason we see the size of a pointer to be 4 bytes in 32 bit machine and 8 bytes in a 64 bit machine

User · Answer

The reason the size of your pointer is 4 bytes is because you are compiling for a 32-bit architecture   As FryGuy pointed out  on a 64-bit architecture you would see 8

User · Answer

In addition to the 16 32 64 bit differences even odder things can occur   There have been machines where sizeof int    will be one value  probably 4 but where sizeof char    is larger   Machines that naturally address words instead of bytes have to  augment  character pointers to specify what portion of the word you really want in order to properly implement the C C   standard   This is now very unusual as hardware designers have learned the value of byte addressability

User · Answer

In addition to the 16 32 64 bit differences even odder things can occur   There have been machines where sizeof int    will be one value  probably 4 but where sizeof char    is larger   Machines that naturally address words instead of bytes have to  augment  character pointers to specify what portion of the word you really want in order to properly implement the C C   standard   This is now very unusual as hardware designers have learned the value of byte addressability

User · Answer

The guarantee you get is that sizeof char     1   There are no other guarantees  including no guarantee that sizeof int       sizeof double      In practice  pointers will be size 2 on a 16-bit system  if you can find one   4 on a 32-bit system  and 8 on a 64-bit system  but there s nothing to be gained in relying on a given size

User · Answer

No  the size of a pointer may vary depending on the architecture  There are numerous exceptions

User · Answer

From what I recall  it s based on the size of a memory address  So on a system with a 32-bit address scheme  sizeof will return 4  since that s 4 bytes

User · Answer

Even on a plain x86 32 bit platform  you can get a variety of pointer sizes  try this out for an example   struct A      struct B   virtual public A      struct C      struct D   public A  public C      int main         cout  lt  lt   A    lt  lt  sizeof void  A         lt  lt  endl      cout  lt  lt   B    lt  lt  sizeof void  B         lt  lt  endl      cout  lt  lt   D    lt  lt  sizeof void  D         lt  lt  endl      Under Visual C   2008  I get 4  12 and 8 for the sizes of the pointers-to-member-function   Raymond Chen talked about this here

User · Answer

Size of pointer and int is 2 bytes in Turbo C compiler on windows 32 bit machine    So size of pointer is compiler specific  But generally most of the compilers are implemented to support 4 byte pointer variable in 32 bit and 8 byte pointer variable in 64 bit machine     So size of pointer is not same in all machines

User · Answer

In addition to what people have said about 64-bit  or whatever  systems  there are other kinds of pointer than pointer-to-object    A pointer-to-member might be almost any size  depending how they re implemented by your compiler  they aren t necessarily even all the same size  Try a pointer-to-member of a POD class  and then a pointer-to-member inherited from one of the base classes of a class with multiple bases  What fun

User · Answer

Even on a plain x86 32 bit platform  you can get a variety of pointer sizes  try this out for an example   struct A      struct B   virtual public A      struct C      struct D   public A  public C      int main         cout  lt  lt   A    lt  lt  sizeof void  A         lt  lt  endl      cout  lt  lt   B    lt  lt  sizeof void  B         lt  lt  endl      cout  lt  lt   D    lt  lt  sizeof void  D         lt  lt  endl      Under Visual C   2008  I get 4  12 and 8 for the sizes of the pointers-to-member-function   Raymond Chen talked about this here

User · Answer

No  the size of a pointer may vary depending on the architecture  There are numerous exceptions

User · Answer

Just another exception to the already posted list  On 32-bit platforms  pointers can take 6  not 4  bytes    include  lt stdio h gt   include  lt stdlib h gt   int main         char far  ptr     note that this is a far pointer     printf    d n   sizeof  ptr        return EXIT SUCCESS      If you compile this program with Open Watcom and run it  you ll get 6  because far pointers that it supports consist of 32-bit offset and 16-bit segment values

User · Answer

The reason the size of your pointer is 4 bytes is because you are compiling for a 32-bit architecture   As FryGuy pointed out  on a 64-bit architecture you would see 8

User · Answer

Technically speaking  the C standard only guarantees that sizeof char     1  and the rest is up to the implementation  But on modern x86 architectures  e g  Intel AMD chips  it s fairly predictable   You ve probably heard processors described as being 16-bit  32-bit  64-bit  etc  This usually means that the processor uses N-bits for integers  Since pointers store memory addresses  and memory addresses are integers  this effectively tells you how many bits are going to be used for pointers  sizeof is usually measured in bytes  so code compiled for 32-bit processors will report the size of pointers to be 4  32 bits   8 bits per byte   and code for 64-bit processors will report the size of pointers to be 8  64 bits   8 bits per byte   This is where the limitation of 4GB of RAM for 32-bit processors comes from -- if each memory address corresponds to a byte  to address more memory you need integers larger than 32-bits

User · Answer

The guarantee you get is that sizeof char     1   There are no other guarantees  including no guarantee that sizeof int       sizeof double      In practice  pointers will be size 2 on a 16-bit system  if you can find one   4 on a 32-bit system  and 8 on a 64-bit system  but there s nothing to be gained in relying on a given size

User · Answer

Technically speaking  the C standard only guarantees that sizeof char     1  and the rest is up to the implementation  But on modern x86 architectures  e g  Intel AMD chips  it s fairly predictable   You ve probably heard processors described as being 16-bit  32-bit  64-bit  etc  This usually means that the processor uses N-bits for integers  Since pointers store memory addresses  and memory addresses are integers  this effectively tells you how many bits are going to be used for pointers  sizeof is usually measured in bytes  so code compiled for 32-bit processors will report the size of pointers to be 4  32 bits   8 bits per byte   and code for 64-bit processors will report the size of pointers to be 8  64 bits   8 bits per byte   This is where the limitation of 4GB of RAM for 32-bit processors comes from -- if each memory address corresponds to a byte  to address more memory you need integers larger than 32-bits

User · Answer

In addition to what people have said about 64-bit  or whatever  systems  there are other kinds of pointer than pointer-to-object    A pointer-to-member might be almost any size  depending how they re implemented by your compiler  they aren t necessarily even all the same size  Try a pointer-to-member of a POD class  and then a pointer-to-member inherited from one of the base classes of a class with multiple bases  What fun

User · Answer

Technically speaking  the C standard only guarantees that sizeof char     1  and the rest is up to the implementation  But on modern x86 architectures  e g  Intel AMD chips  it s fairly predictable   You ve probably heard processors described as being 16-bit  32-bit  64-bit  etc  This usually means that the processor uses N-bits for integers  Since pointers store memory addresses  and memory addresses are integers  this effectively tells you how many bits are going to be used for pointers  sizeof is usually measured in bytes  so code compiled for 32-bit processors will report the size of pointers to be 4  32 bits   8 bits per byte   and code for 64-bit processors will report the size of pointers to be 8  64 bits   8 bits per byte   This is where the limitation of 4GB of RAM for 32-bit processors comes from -- if each memory address corresponds to a byte  to address more memory you need integers larger than 32-bits

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