C 中的类型提升 [英] type promotion in C

问题描述

我对以下代码感到很困惑:

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

    return 0;
}

返回:

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

似乎表达式 (a - b) 的类型是 uint32_t.我不明白为什么，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.

谁能给我解释一下?

推荐答案

C 标准非常清楚地解释了这一点(§6.5.6 Additive Operators):

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.

If both operands have arithmetic type, the usual arithmetic conversions are performed on them.

(第 6.3.1.8 节常用算术转换):

(§6.3.1.8 Usual Arithmetic Conversions):

...整数提升对两个操作数执行.

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

(第 6.3.1.1 节布尔值、字符和整数):

(§6.3.1.1 Boolean, characters, and integers):

如果一个int可以表示原始类型的所有值，则将该值转换为int；...这些被称为整数提升.所有其他类型都不会因整数提升而改变.

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可以表示你平台上uint16_t的所有值，a和b被转换为int 在执行减法之前.结果的类型为 int，并作为 int 传递给 printf.您已使用 int 参数指定了 %u 格式化程序；严格来说，这会调用未定义的行为，但在您的平台上，int 参数被解释为它的二进制补码表示，并被打印出来.

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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