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C89和C99汇编后,C方案在日常工作中的表现不同?
原标题:What C program behaves differently in run-time when compiled with C89 and C99?

我发现以下幻灯(我想在Wikipedia)在C++评论得到确认时,会产生不同时间:

int a = 4 //* This is a comment, but where does it end? */ 2
  ;

但直到现在,这只是唯一的(被排除在外的人)。

我不想使用<代码>__STDC__等加以区别,也不希望在C89不汇编的方案中加以区别。

是否有其他方案/工具,其运行时间不同于C89,而不是C99?

最佳回答

本方案将印刷关于符合《公约》第89条执行情况的<编码>0.000000<>/code>和关于符合《公约》第99条执行情况的<代码>1.000000<>:>:

#include <stdio.h>
#include <stdlib.h>

int main()
{
    double d = strtod("0x1", NULL);
    printf("%f
", d);
    return 0;
}
问题回答

两个例子:

  • C99有-3/2 被定义为Behaviour(即拖到零)。

  • C99有-1<<1,作为未定义的Behaviour(但不是C89)。

Also, in the past I ve run into problems with 64-bit enums, such as enum {mask = 1ULL << 32}, but I don t recall if the compiler was silent, or just quietly did the wrong thing.

愤怒司可产生不同的结果,取决于你使用哪种分类方法89。

Does either ANSI C or ISO C specify what -5 % 10 should be?





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