C语言中的类型提升 [英] type promotion in C

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;
}

返回:

$ 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):

如果两个操作数都具有算术类型，则对它们执行通常的算术转换.

(第6.3.1.8节常规算术转换):

(§6.3.1.8 Usual Arithmetic Conversions):

... 整数提升在两个操作数上执行.

... the integer promotions are performed on both operands.

(第6.3.1.1节布尔，字符和整数):

(§6.3.1.1 Boolean, characters, and integers):

如果int可以代表原始类型的所有值，则该值将转换为int；否则，该值将转换为int. ...这些称为整数促销.整数促销未更改所有其他类型.

If an int can represent all values of the original type, the value is converted to an int; ... These are called the integer promotions. All other types are unchanged by the integer promotions.

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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