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)，正如我所期望的那样。毕竟，我们这里只讨论个位数，不是吗？那么为什么在使用%d而不是时会出现以下错误%lu？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

在 C 中，表达式的类型sizeof是size_t。

printf要使用的说明符是%zu. 例子：

printf("The 'int' datatype is \t\t %zu bytes\n", sizeof(int));

从 1999 年版本的 C 标准开始就可以使用它了。

Answer 2

您正在尝试打印sizeofoperator的返回值，它通常是 type size_t。

在您的情况下，它似乎size_t是 a typedefof long unsigned int，因此它要求%lu使用兼容的格式说明符。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，%zu在基于 C99 和前向标准的编译器上使用,是安全的。