c - 为什么 char 字符串中没有取消引用运算符？

Question

假设我有两个函数:

void function1(int *ptr) {
     printf("%d", *ptr);
}

和

void function2(char *str) {
     printf("%s", str);
}

当 str 之前没有遵从运算符时，为什么 function2 可以工作？在 str 中只有 address 指向的不是我想的 value。

Answer 1

Why does function2 work, when there is no deference operator before str

因为标准定义的 %s 需要一个 char *，即一个地址。

7.21.6.1

If no l length modifier is present, the argument shall be a pointer to the initial element of an array of character type. Characters from the array are written up to (but not including) the terminating null character. If the precision is specified, no more than that many bytes are written. If the precision is not specified or is greater than the size of the array, the array shall contain a null character.