如何打印 uint32_t 和 uint16_t 变量的值? [英] How do I print uint32_t and uint16_t variables' value?

问题描述

我正在尝试打印 uint16_t 和 uint32_t 值，但它没有提供所需的输出.

I am trying to print an uint16_t and uint32_t value, but it is not giving the desired output.

#include <stdio.h>
#include <netinet/in.h>

int main()
{
    uint32_t a = 12, a1;
    uint16_t b = 1, b1;
    a1 = htonl(a);
    printf("%d---------%d", a1);
    b1 = htons(b);
    printf("
%d-----%d", b, b1);
    return 0;
}

我也用过

 printf("%"PRIu32, a);

显示错误.

如何打印这些值以及所需的输出是什么?

How do I print these values and what will be the desired output?

推荐答案

如果你想要 intN_t 的所有那些漂亮的新格式说明符，你需要包含 inttypes.h类型和它们的兄弟，这是正确的(即，可移植的)方法，前提是您的编译器符合 C99.您不应该使用像 %d 或 %u 这样的标准尺寸，以防尺寸与您的想法不同.

You need to include inttypes.h if you want all those nifty new format specifiers for the intN_t types and their brethren, and that is the correct (ie, portable) way to do it, provided your compiler complies with C99. You shouldn't use the standard ones like %d or %u in case the sizes are different to what you think.

它包括 stdint.h 并扩展了很多其他东西，例如可用于 printf/scanf 调用系列的宏.这在 ISO C99 标准的第 7.8 节中有介绍.

It includes stdint.h and extends it with quite a few other things, such as the macros that can be used for the printf/scanf family of calls. This is covered in section 7.8 of the ISO C99 standard.

例如下面的程序:

#include <stdio.h>
#include <inttypes.h>
int main (void) {
    uint32_t a=1234;
    uint16_t b=5678;
    printf("%" PRIu32 "
",a);
    printf("%" PRIu16 "
",b);
    return 0;
}

输出:

1234
5678

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