CODEWITHSUNDEEP
c++gcc-warning
Souvik Pal
Souvik Pal

Reputation: 63

warning: format ‘%u’ expects argument of type ‘unsigned int’, but argument 2 has type ‘int*’ [-Wformat=]

While I'm compiling this simple program in GCC compiler I'm getting this error:- warning: format ‘%u’ expects argument of type ‘unsigned int’, but argument 2 has type ‘int’ [-Wformat=]*

#include <stdio.h>
int main()
{
    printf("Pointer\n");
    printf("*******\n\n");
    int i=3;
    printf("Address of Variable i : %u",&i);
    printf("Value stored in Variable i : %d\n",i);
    printf("Value stored in Variable i : %d\n",*(&i));
    return 0;
}

Upvotes: 2

Views: 12917

Answers (1)

Olaf Dietsche
Olaf Dietsche

Reputation: 74118

Even without any extra option, gcc (9.3.0) shows a detailed warning message

a.cpp: In function 'int main()':
a.cpp:7:38: warning: format '%u' expects argument of type 'unsigned int', but argument 2 has type 'int*' [-Wformat=]
    7 |     printf("Address of Variable i : %u",&i);
      |                                     ~^  ~~
      |                                      |  |
      |                                      |  int*
      |                                      unsigned int
      |                                     %n

So there's a mismatch between the format %u (unsigned int) and the argument &i (pointer).

Looking at printf Conversion specifier

u converts an unsigned integer into decimal representation dddd.
...
p writes an implementation defined character sequence defining a pointer.

The proper format in this case would be %p for a pointer argument &i

printf("Address of Variable i : %p", &i);

Upvotes: 2

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