Converting an int into a 4 byte char array C

Question

Hey  I m looking to convert a int that is inputed by the user into 4 bytes  that I am assigning to a character array  How can this be done   Example   Convert a user inputs of 175 to  00000000 00000000 00000000 10101111    Issue with all of the answers so far  converting 255 should result in 0 0 0 ff although it prints out as  0 0 0 ffffffff  unsigned int value   255      buffer 0     value  gt  gt  24   amp  0xFF  buffer 1     value  gt  gt  16   amp  0xFF  buffer 2     value  gt  gt  8   amp  0xFF  buffer 3    value  amp  0xFF   union       unsigned int integer      unsigned char byte 4     temp32bitint   temp32bitint integer   value  buffer 8    temp32bitint byte 3   buffer 9    temp32bitint byte 2   buffer 10    temp32bitint byte 1   buffer 11    temp32bitint byte 0     both result in 0 0 0 ffffffff instead of 0 0 0 ff  Just another example is 175 as the input prints out as 0  0  0  ffffffaf when it should just be 0  0  0  af

User · Answer

int a   1  char   c    char    amp a     In C   should be intermediate cst to void

User · Answer

The problem is arising as unsigned char is a 4 byte number not a 1 byte number as many think  so change it to  union   unsigned int integer  char byte 4     temp32bitint    and cast while printing  to prevent promoting to  int   which C does by default   printf   u   u  n    unsigned char Buffer 0    unsigned char Buffer 1

User · Answer

An int is equivalent to uint32 t and char to uint8 t   I ll show how I resolved client-server communication  sending the actual time  4 bytes  formatted in Unix epoch  in a 1-bit array  and then re-built it in the other side   Note  the protocol was to send 1024 bytes    Client side  uint8 t message 1024   uint32 t t   time NULL    uint8 t watch 4      t  amp  255   t  gt  gt  8   amp  255   t  gt  gt  16   amp  255   t  gt  gt   24   amp  255     message 0    watch 0   message 1    watch 1   message 2    watch 2   message 3    watch 3   send socket  message  1024  0    Server side  uint8 t res 1024   uint32 t date   recv socket  res  1024  0    date   res 0     res 1   lt  lt  8     res 2   lt  lt  16     res 3   lt  lt  24    printf  Received message from client  d sent at  d n   socket  date      Hope it helps

User · Answer

Why would you need an intermediate cast to void   in C   Because cpp doesn t allow direct conversion between pointers  you need to use reinterpret cast or casting to void  does the thing

User · Answer

In your question  you stated that you want to convert a user input of 175 to 00000000 00000000 00000000 10101111  which is big endian byte ordering  also known as network byte order   A mostly portable way to convert your unsigned integer to a big endian unsigned char array  as you suggested from that  175  example you gave  would be to use C s htonl   function  defined in the header  lt arpa inet h gt  on Linux systems  to convert your unsigned int to big endian byte order  then use memcpy    defined in the header  lt string h gt  for C   lt cstring gt  for C    to copy the bytes into your char  or unsigned char  array   The htonl   function takes in an unsigned 32-bit integer as an argument  in contrast to htons    which takes in an unsigned 16-bit integer  and converts it to network byte order from the host byte order  hence the acronym  Host TO Network Long  versus Host TO Network Short for htons   returning the result as an unsigned 32-bit integer  The purpose of this family of functions is to ensure that all network communications occur in big endian byte order  so that all machines can communicate with each other over a socket without byte order issues   As an aside  for big-endian machines  the htonl    htons    ntohl   and ntohs   functions are generally compiled to just be a  no op   because the bytes do not need to be flipped around before they are sent over or received from a socket since they re already in the proper byte order   Here s the code    include  lt stdio h gt   include  lt arpa inet h gt   include  lt string h gt   int main         unsigned int number   175       unsigned int number2   htonl number       char numberStr 4       memcpy numberStr   amp number2  4        printf   x  x  x  x n   numberStr 0   numberStr 1   numberStr 2   numberStr 3         return 0      Note that  as caf said  you have to print the characters as unsigned characters using printf s  x format specifier   The above code prints 0 0 0 af on my machine  an x86 64 machine  which uses little endian byte ordering   which is hex for 175

User · Answer

You can try   void CopyInt int value  char  buffer      memcpy buffer   void  value  sizeof int

User · Answer

You can simply use memcpy as follows   unsigned int value   255  char bytes 4     0  0  0  0   memcpy bytes   amp value  4

User · Answer

The portable way to do this  ensuring that you get 0x00 0x00 0x00 0xaf everywhere  is to use shifts   unsigned char bytes 4   unsigned long n   175   bytes 0     n  gt  gt  24   amp  0xFF  bytes 1     n  gt  gt  16   amp  0xFF  bytes 2     n  gt  gt  8   amp  0xFF  bytes 3    n  amp  0xFF    The methods using unions and memcpy   will get a different result on different machines     The issue you are having is with the printing rather than the conversion   I presume you are using char rather than unsigned char  and you are using a line like this to print it   printf   x  x  x  x n   bytes 0   bytes 1   bytes 2   bytes 3      When any types narrower than int are passed to printf  they are promoted to int  or unsigned int  if int cannot hold all the values of the original type    If char is signed on your platform  then 0xff likely does not fit into the range of that type  and it is being set to -1 instead  which has the representation 0xff on a 2s-complement machine    -1 is promoted to an int  and has the representation 0xffffffff as an int on your machine  and that is what you see   Your solution is to either actually use unsigned char  or else cast to unsigned char in the printf statement   printf   x  x  x  x n    unsigned char bytes 0                            unsigned char bytes 1                            unsigned char bytes 2                            unsigned char bytes 3

User · Answer

The issue with the conversion  the reason it s giving you a ffffff at the end  is because your hex integer  that you are using the  amp  binary operator with  is interpreted as being signed  Cast it to an unsigned integer  and you ll be fine

User · Answer

Do you want to address the individual bytes of a 32-bit int  One possible method is a union   union       unsigned int integer      unsigned char byte 4     foo   int main         foo integer   123456789      printf   u  u  u  u n   foo byte 3   foo byte 2   foo byte 1   foo byte 0        Note  corrected the printf to reflect unsigned values

[c] Converting an int into a 4 byte char array (C)

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