C语言 “%d”需要“int”类型的参数，但参数 2 的类型为“long unsigned int”[-Wformat=]

'%d' expects argument of type 'int', but argument 2 has type 'long unsigned int' [-Wformat=]

提问by LuizGsa13

I keep getting compile warnings but I don't know how to fix it:

我不断收到编译警告，但我不知道如何修复它：

'%d' expects argument of type 'int', but argument 2 has type 'long unsigned int' [

The program runs fine but I still get the compile warnings:

程序运行良好，但我仍然收到编译警告：

/* Sizeof.c--Program to tell byte size of the C variable */
#include <stdio.h>

int main(void) {
    printf("\nA Char is %d bytes", sizeof( char ));
    printf("\nAn int is %d bytes", sizeof( int ));
    printf("\nA short is %d bytes", sizeof( short ));
    printf("\nA long is %d bytes", sizeof( long ));
    printf("\nA long long is %d bytes\n", sizeof( long long ));
    printf("\nAn unsigned Char is %d bytes", sizeof( unsigned char ));
    printf("\nAn unsigned int is %d bytes", sizeof( unsigned int));
    printf("\nAn unsigned short is %d bytes", sizeof( unsigned short ));
    printf("\nAn unsigned long is %d bytes", sizeof( unsigned long ));
    printf("\nAn unsigned long long is %d bytes\n",
            sizeof( unsigned long long ));
    printf("\nfloat is %d bytes", sizeof( float ));
    printf("\nA double is %d bytes\n", sizeof( double ));
    printf("\nA long double is %d bytes\n", sizeof( long double ));

    return 0;

}

回答by Shafik Yaghmour

sizeofreturns size_tyou need to use %zufor the format string instead of %d. The type of unsigned integerof size_tcan vary (depending on platform) and may not be long unsigned inteverywhere, which is covered in the draft C99 standard section 6.5.3.4The sizeof operatorparagraph 4:

sizeof返回size_t您需要%zu用于格式字符串而不是%d. 所述类型的无符号整数的size_t可以变化（取决于平台），也有可能长UNSIGNED INT无处不在，其在C99草案标准部覆盖6.5.3.4sizeof运算符段4：

The value of the result is implementation-defined, and its type (an unsigned integer type) is size_t, defined in (and other headers).

结果的值是实现定义的，它的类型（无符号整数类型）是 size_t，定义在（和其他头文件）中。

Also note that using the wrong format specifier for printfis undefined behavior, which is covered in section 7.19.6.1The fprintf function, which also covers printfwith respect to format specifiers says:

另请注意，使用错误的格式说明符printf是未定义的行为，这在7.19.6.1The fprintf function部分中进行了介绍，该部分还涵盖printf了关于格式说明符的内容：

If a conversion specification is invalid, the behavior is undefined.²⁴⁸⁾If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined.

如果转换规范无效，则行为未定义。²⁴⁸⁾如果任何参数不是相应转换规范的正确类型，则行为未定义。

Update

更新

Visual Studiodoes not support the zformat specifier:

Visual Studio不支持z格式说明符：

The hh, j, z, and t length prefixes are not supported.

不支持 hh、j、z 和 t 长度前缀。

the correct format specifier in this case would be %Iu.

在这种情况下，正确的格式说明符是%Iu.

回答by Fiddling Bits

The compiler is warning you that you may suffer a loss of precision. That is, the format specifier that you're using to print a sizeof, %d, is not capable of printing the full range of size_t. Change %dto %zuand your warning will go away.

编译器警告您可能会损失精度。也就是说，格式说明您使用打印sizeof，%d是不是能够印刷的全方位的size_t。更改%d为%zu，您的警告将消失。

回答by Suraj

i had the same problem in Linux. the same program runs without error in windows (means '%d' worked without error), but for linux i had to replace all the '%d' with'%lu' to run the program.

我在 Linux 中遇到了同样的问题。相同的程序在 Windows 中运行没有错误（意味着 '%d' 工作没有错误），但对于 linux，我必须用 '%lu' 替换所有的 '%d' 才能运行该程序。