How does assigning a string to int and passing that int to printf prints the string properly?

Question

Why does this work? (i.e how is passing int to printf() results in printing the string)

#include<stdio.h>

int main() {
    int n="String";
    printf("%s",n);
    return 0;
}

warning: initialization makes integer from pointer without a cast [enabled by default]
int n="String";
warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int’ [-Wformat=]
printf("%s",n);

Output:String

compiler:gcc 4.8.5

Sourav Ghosh Sourav Ghosh · Accepted Answer · 2016-08-12T12:12:16

In your code,

  int n="String";  //conversion of pointer to integer

is highly-implementation dependent ^{note 1}

and

 printf("%s",n);   //passing incompatible type of argument

invokes undefined behavior. ^{note 2}Don't do that.

Moral of the story: The warnings are there for a reason, pay heed to them.

Note 1:

Quoting C11, chapter §6.3.2.3

Any pointer type may be converted to an integer type. Except as previously specified, the result is implementation-defined. If the result cannot be represented in the integer type, the behavior is undefined. [....]

Note 2:

chapter §7.21.6.1

[....] If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined.

and for the type of argument for %s format specifier with printf()

s If no l length modifier is present, the argument shall be a pointer to the initial element of an array of character type. [...]

How does assigning a string to int and passing that int to printf prints the string properly?

4 Answers