I'm testing this code ,but why no error at all?
#include <stdio.h> int main() { int a = 1025; int *p; p = &a; // now I declare a char variable , char *p0; p0 = (char*) p; // type casting printf("", sizeof(char)); // is %d correct here ????? printf("Address = %d, value = %d\n", p0, *p0); } My question : Is %d correct here? since %d is integer not for character, why no error at all?
In your case
p0 = (char*) p; is valid, because char * can be used to access any other types. Related, quoting C11, chapter §6.3.2.3
[...] When a pointer to an object is converted to a pointer to a character type, the result points to the lowest addressed byte of the object. Successive increments of the result, up to the size of the object, yield pointers to the remaining bytes of the object.
However, in case of
printf("Address = %d, value = %d\n", p0, *p0); causes undefined behavior, as you're passing a pointer (p0) to %d (Look at the first "pair" of conversion specifier and corresponding argument). You should use %p and cast the argument to void *, something like
printf("Address = %p, value = %d\n", (void *)p0, *p0); Then, to answer
No error at all, why?
because the issue is not with the syntax or any constraint violation which the compiler is supposed to complain about. It's purely mis-use of given power. :)
It's undefined behaviour because p0 is a char*, the code printf("Address = %d, value = %d\n", p0, *p0) and %d is not a valid conversion specifier for character pointers.
constraint violation...can you please elaborate?
%d is not a valid conversion specifier for character pointers it is not for any pointer, as a matter of fact.
%dwhere you are passing an address: you should use%p.printf("Address = %p, value = %d\n", (void *)p0, *p0);For the second onecharvalue is just promoted toint.gcc -Wall -Wextra -pedantic-errors test.c -o test -std=c11 -g, you havewarning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘char *’ [-Wformat=] printf("Address = %d, value = %d\n", p0, *p0);