C语言中的类型提升 | C语言中的类型提升

本文介绍了C语言中的类型提升的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我对以下代码感到非常困惑:

I am quite confused by the following code:

#include <stdio.h>
#include <stdint.h>

int main(int argc, char ** argv)
{
    uint16_t a = 413;
    uint16_t b = 64948;

    fprintf(stdout, "%u\n", (a - b));
    fprintf(stdout, "%u\n", ((uint16_t) (a - b)));

    return 0;
}

$ gcc -Wall test.c -o test
$ ./test
4294902761
1001
$

似乎表达式(a-b)的类型为uint32_t.我不理解为什么，因为两个运算符都是uint16_t.

It seems that expression (a - b) has type uint32_t.I don't uderstand why since both operators are uint16_t.

有人可以向我解释吗?

推荐答案

C标准对此进行了很清楚的解释(第6.5.6节加法运算符"):

The C standard explains this quite clearly (§6.5.6 Additive Operators):

(第6.3.1.8节常规算术转换):

(§6.3.1.8 Usual Arithmetic Conversions):

(第6.3.1.1节布尔，字符和整数):

(§6.3.1.1 Boolean, characters, and integers):

由于int可以表示平台上所有uint16_t的值，因此在执行减法之前，a和b会转换为int.结果的类型为int，并作为int传递给printf.您已经使用int参数指定了%u格式化程序；严格来说，这会引起未定义的行为，但是在您的平台上，int参数被解释为二进制补码表示，并且会被打印出来.

Since int can represent all values of uint16_t on your platform, a and b are converted to int before the subtraction is performed. The result has type int, and is passed to printf as an int. You have specified the %u formatter with an int argument; strictly speaking this invokes undefined behavior, but on your platform the int argument is interpreted as it's twos-complement representation, and that is printed.

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