C 中 sizeof() 返回值的正确格式说明符

Question

我有以下代码:

#include<stdio.h>

int main()
{
    printf("The 'int' datatype is \t\t %lu bytes\n", sizeof(int));
    printf("The 'unsigned int' data type is\t %lu bytes\n", sizeof(unsigned int));
    printf("The 'short int' data type is\t %lu bytes\n", sizeof(short int));
    printf("The 'long int' data type is\t %lu bytes\n", sizeof(long int));
    printf("The 'long long int' data type is %lu bytes\n", sizeof(long long int));
    printf("The 'float' data type is\t %lu bytes\n", sizeof(float));
    printf("The 'char' data type is\t\t %lu bytes\n", sizeof(char));
}

哪些输出:

The 'int' datatype is        4 bytes
The 'unsigned int' data type is  4 bytes
The 'short int' data type is     2 bytes
The 'long int' data type is  8 bytes
The 'long long int' data type is 8 bytes
The 'float' data type is     4 bytes
The 'char' data type is      1 bytes

但这就是问题，编译器要求我使用 %lu (long unsigned int) 而不是 %d (int)，正如我所期望的。毕竟，我们这里只讨论个位数，不是吗？那么为什么在使用 %d 而不是 %lu 时会出现以下错误？这与我使用 64 位系统(Ubuntu 14.10)有关吗？

helloworld.c: In function ‘main’:
helloworld.c:5:5: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘long unsigned int’ [-Wformat=]
     printf("The 'int' datatype is \t\t %d bytes\n", sizeof(int));
     ^
helloworld.c:6:5: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘long unsigned int’ [-Wformat=]
     printf("The 'unsigned int' data type is\t %d bytes\n", sizeof(unsigned int));
     ^
helloworld.c:7:5: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘long unsigned int’ [-Wformat=]
     printf("The 'short int' data type is\t %d bytes\n", sizeof(short int));
     ^
helloworld.c:8:5: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘long unsigned int’ [-Wformat=]
     printf("The 'long int' data type is\t %d bytes\n", sizeof(long int));
     ^
helloworld.c:9:5: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘long unsigned int’ [-Wformat=]
     printf("The 'long long int' data type is %d bytes\n", sizeof(long long int));
     ^
helloworld.c:10:5: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘long unsigned int’ [-Wformat=]
     printf("The 'float' data type is\t %d bytes\n", sizeof(float));
     ^
helloworld.c:11:5: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘long unsigned int’ [-Wformat=]
     printf("The 'char' data type is\t\t %d bytes\n", sizeof(char));
     ^
Compilation finished successfully.

Answer 1

您正在尝试打印 sizeof operator 的返回值，该值通常是 size_t 类型。

看来，在您的情况下， size_t 是 typedef 的 long unsigned int ，因此它要求使用它的兼容格式说明符 %lu 。这里的返回值无关紧要，您的问题在于类型不匹配。

注意:要获得可移植的代码，在基于 C99 和转发标准的编译器上使用 %zu 是安全的。