sizeof 和 strlen 的printf 问题

#include <stdio.h>
#include <string.h>

int main(void){
    char temp[3];
    scanf("%s", temp);
    printf("%d
", sizeof(temp));
    printf("%d
", strlen(temp));
    printf("%s
", temp);
}

编译时出错：

test_scanf.c: In function ‘main’:
test_scanf.c:7:2: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘long unsigned int’ [-Wformat=]
  printf("%d
", sizeof(temp));
  ^
test_scanf.c:8:2: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘size_t’ [-Wformat=]
  printf("%d
", strlen(temp));
  ^

sizeof() 和 strlen() 返回值类型为size_t 

 In C99's *printf, z is a length modifier like h or l, not a conversion specifier like d or u. 

将代码改为：

printf("%zu
", sizeof(temp));
printf("%zu
", strlen(temp));

 另外：

strlen 计算到 出现为止。

sizeof 保证能容纳实现所建立的最大对象的字节大小。