error invalid type argument of unary have int

Question

I have a C Program    include  lt stdio h gt  int main      int b   10                assign the integer 10 to variable  b     int  a                    declare a pointer to an integer  a     a  int    amp b               Get the memory location of variable  b  cast it                             to an int pointer and assign it to pointer  a     int  c                    declare a pointer to an integer  c     c  int    amp a               Get the memory location of variable  a  which is                             a pointer to  b    Cast that to an int pointer                              and assign it to pointer  c      printf   d     c          ERROR HAPPENS HERE       return 0          Compiler produces an error    error  invalid type argument of    unary       have    int       Can someone explain what this error means

User · Answer

Barebones C program to produce the above error    include  lt iostream gt  using namespace std  int main        char  p       p    c        cout  lt  lt   p 0           error  invalid type argument of  unary          peeking too deeply into p  that s a paddlin       cout  lt  lt    p            error  invalid type argument of  unary          peeking too deeply into p  you better believe that s a paddlin      ELI5   The master puts a shiny round stone inside a small box and gives it to a student   The master says   Open the box and remove the stone    The student does so     Then the master says   Now open the stone and remove the stone    The student said   I can t open a stone      The student was then enlightened

User · Answer

I have reformatted your code   The error was situated in this line    printf   d      c      To fix it  change to    printf   d     c      The   retrieves the value from an address  The    retrieves the value  an address in this case  of an other value from an address   In addition  the    was optional    include  lt stdio h gt   int main void        int b   10       int  a   NULL      int  c   NULL       a    amp b      c    amp a       printf   d    c        return 0       EDIT    The line    c    amp a    must be replaced by    c   a    It means that the value of the pointer  c  equals the value of the pointer  a   So   c  and  a  points to the same address   b    The output is    10   EDIT 2   If you want to use a double       include  lt stdio h gt   int main void        int b   10       int  a   NULL      int   c   NULL       a    amp b      c    amp a       printf   d     c        return 0       Output   10

User · Answer

Once you declare the type of a variable  you don t need to cast it to that same type  So you can write a  amp b    Finally  you declared c incorrectly  Since you assign it to be the address of a  where a is a pointer to int  you must declare it to be a pointer to a pointer to int    include  lt stdio h gt  int main void        int b 10      int  a  amp b      int   c  amp a      printf   d     c       return 0

User · Answer

Since c is holding the address of an integer pointer  its type should be int     int   c  c    amp a    The entire program becomes    include  lt stdio h gt                                                                int main        int b 10      int  a      a  amp b      int   c      c  amp a      printf   d     c        successfully prints 10     return 0

[c] error: invalid type argument of ‘unary *’ (have ‘int’)

Examples related to c

Examples related to pointers