c - 将字符串分配给 int 并将该 int 传递给 printf 如何正确打印字符串？

Question

为什么会这样？ (即如何将 int 传递给 printf() 结果打印字符串)

#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);

输出:字符串

编译器:gcc 4.8.5

Answer 1

在你的代码中，

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

高度依赖于^注释1

和

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

调用 undefined behavior . ^{注释 2} 不要那样做。

故事的寓意:警告是有原因的，请注意它们。

---

注1:

引用 C11，章节 §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. [....]

注2:

§7.21.6.1 章

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

%s 格式说明符的参数类型与 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. [...]