为什么这两个程序的行为不同的ANSI C？ [英] Why did these two program behave differently in ANSI C?

问题描述

可能重复：结果
  谜语（C语言）

1

main()
{

 if(-1<(unsigned char)1)
     printf("-1 is less than (unsigned char)1:ANSI semantics");
 else
     printf("-1 NOT less than (unsigned char)1:K&R semantics");
}

2

int array[] = {23,41,12,24,52,11};
#define TOTAL_ELEMENTS (sizeof(array)/sizeof(array[0]))
main()
{
    int d = -1,x;
    if(d<=TOTAL_ELEMENTS -2)
        x = array[d+1];
}

第一个转换unsigned char型1到符号变量在ANSI C，
而第二个程序转换d可一个unsigned int，使
条件前pression在ANSI C.返回false
为什么他们不同的表现？

The first convert unsigned char 1 to a signed variable in ANSI C, while the second program convert d to an unsigned int that makes the condition expression return false in ANSI C. Why did they behave differently?

推荐答案

有关第一个右手边是一个无符号的字符，以及所有无符号的字符值放入一个符号整数，所以它转换为符号整数。

For the first one the right-hand side is an unsigned char, and all unsigned char values fit into a signed int, so it is converted to signed int.

有关第二个右手边是一个unsigned int，因此左侧是由符号int转换为unsigned int类型。

For the second one the right-hand side is an unsigned int, so the left-hand side is converted from signed int to unsigned int.

又见<一个href=\"https://www.securecoding.cert.org/confluence/display/sec$c$c/INT02-C.+Understand+integer+conversion+rules\"相对=nofollow>整数转换这CERT文档。

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