I am currently doing pointers. I have been programming for a long time, but not in C/C . With that being said, my pointer knowledge is abysmal. Currently, I am following a guide on YouTube and he prints the code below.

int main() {

int a = 5;

int *p;

p = &a;

printf("%d\n", p);
}

This prints successfully for him, and he sees a memory location. For me, I see the error

warning: format '%d' expects argument of type 'int', but argument 2 has type 'int *'

From this, I expect I need to put an & in front of the p to make it print the value. But then I receive this error,

int main() {

int a = 5;

int *p;

p = &a;

printf("%d\n", &p);
}

'int', but argument 2 has type 'int **'

Where is the hole in my knowledge? Any key tips or strategies when working with this, I don't know why I find this so abstract. Thanks,

I was expecting the value to print as expected, but instead am greeted with the error.

CodePudding user response：

%d is the wrong format specifier for pointers. That may work on a more lenient or noncompliant implementation, but you should use %p to print pointer values.

CodePudding user response：

Warnings are not errors. You're receiving a warning, which is not stopping your program from working, because the print specifier %d (ie printf("%d")) is for displaying integers, and you're giving it a non-integer argument of type int*.

The problem here is not with the argument, it's with the print specifier. Your attempt at a fix just changes the int* to an int**, which still does not match the format specifier %d. Use %p instead, which is the specifier for pointers, and will fix the warning, and print the address in hexadecimal notation.

You could also suppress the warning with a series of explicit casts from int* to int, but integer representations of memory addresses are generally much less used than hexadecimal representations in the first place.

CodePudding user response：

Note that using wrong format specifier in printf() lead to undefined behaviour¹⁾. The correct format specifier to print a pointer is %p format specifier.

Remember, format specifier %p expect that the argument shall be a pointer to void, so you should type cast pointer argument to void *. The correct statement to print pointer p would be:

printf("%p\n", (void *)p);

1. C11#7.21.6.1p9 [emphasis added]

9 If a conversion specification is invalid, the behavior is undefined.282) If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined.