问题描述
我正在尝试打印uint16_t和uint32_t值,但是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 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("\n%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
如果您要为intN_t
类型及其兄弟创建所有漂亮的新格式说明符,并且为正确的(即可移植的)方法,前提是您的编译器符合C99.如果大小与您的想法不同,则不应使用%d
或%u
之类的标准字体.
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
,并用很多其他东西扩展了它,例如可以用于printf/scanf
调用族的宏. ISO C99标准的7.8节对此进行了介绍.
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,以下程序:
#include <stdio.h>
#include <inttypes.h>
int main (void) {
uint32_t a=1234;
uint16_t b=5678;
printf("%" PRIu32 "\n",a);
printf("%" PRIu16 "\n",b);
return 0;
}
输出:
1234
5678
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