当b为零时，为什么a％b会产生SIGFPE？ [英] Why does a%b produce SIGFPE when b is zero?

问题描述

今天，我正在跟踪刚刚编写的某些代码中的浮点异常。找到它花了一些时间，因为它实际上是由整数mod 0引起的。显然，将不会定义mod 0的任何操作，但我认为错误如此令人误解是很奇怪的。在C ++模运算符中，将浮点数用于两个整数是什么？ （我正在使用gcc 4.3.2）

Today I was tracking down a floating point exception in some code I had just written. It took a little while to find because it was actually caused by taking an integer mod zero. Obviously doing anything mod zero is not going to be defined but I thought it was strange that the error was so misleading. What is it within the C++ modulo operator that would use floating point for two integers? (I'm using gcc 4.3.2)

这是一个简单的程序来演示错误。

Here's a simple program to demonstrate the error.

int main()
{
    int a=3,b=0;
    int c=a%b;
    return 0;
}

推荐答案

操作触发 SIGFPE ：

SIG是信号
名称的通用前缀； FPE是
浮点异常的首字母缩写。尽管
SIGFPE不一定涉及
浮点运算，但是没有
的更改方式，而不会破坏
的向后兼容性。

SIG is a common prefix for signal names; FPE is an acronym for floating-point exception. Although SIGFPE does not necessarily involve floating-point arithmetic, there is no way to change its name without breaking backward compatibility.

GDB对此更为清晰，称之为算术异常：

GDB is a bit clearer about this and calls it "Arithmetic exception":

(gdb) run
Starting program: /home/emil/float

Program received signal SIGFPE, Arithmetic exception.
0x0804837d in main () at float.c:4
4           int c=a%b;

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