c中的signed_extension

概述我已经尝试了以下程序并将输出设置为72.我无法理解它如何输出为72. #include <stdio.h>int main(){ int a=0x27ea4d72; char x=a; printf("%x",x); return 0;} 从C标准： 6.3.1.3 Signed and unsigned integers When a value with 我已经尝试了以下程序并将输出设置为72.我无法理解它如何输出为72.

#include <stdio.h>int main(){    int a=0x27ea4d72;    char x=a;    printf("%x",x);    return 0;}

解决方法 从C标准：

6.3.1.3 Signed and unsigned integers

When a value with integer type is converted to another integer
type other than _Bool,if the value can be represented by the new
type,it is unchanged. Otherwise,if the new type is unsigned,the value is converted by
repeatedly adding or subtracting one more than the maximum value that
can be represented in the new type until the
value is in the range of the new type. Otherwise,the new type is signed and the value cannot be
represented in it; the result is
implementation-defined.

转换属于案例#3,因为值已签名.所以结果是实现定义的.

但是,对于代码：

#include <stdio.h>int main(){    unsigned long a = 0x27ea4d72ul;    unsigned char x = a;    printf("%x",x);    return 0;}

转换属于情况#2,在这种情况下,标准基本上表示结果将是0x27ea4d72％(UCHAR_MAX 1).在UCHAR_MAX为255(最常见的情况)的情况下,您将获得0x72.

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