书中要求写一个带有两个整形参数的函数,要求判断是否会发生溢出,书中给出了一段错误代码,问我为什么这么判断是错误的,我只能举出反例,却说不上为什么.代码如下:
#include<stdio.h>
int tadd_ok(int x, int y){
int sum = x + y;
return (sum - x == y) && (sum - y == x);
}我自己尝试着解释了一下, 不知道正不正确
/*
The aim of this function is to judge if the addition of two number
will result overflow. If not return 1 else 0.
*/
int tadd_ok(int x, int y){
int sum = x + y;
return (sum - x == y) && (sum - y == x);
}
/*
The is the reason why this is wrong :
there are two situations that will cause overflow.
assume the int is w-bit-long.
1. two negtive number and get a positive number.
In this case : x + y < -2^(w-1) => since the minium number of int is -2^(w-1)
so in fact sum = (x + y + 2^w) => for example -1 + -8 will get 7 if w = 4.
so when you use sum - x you will get (y + 2^w) % 2^w result y
(when the value is bigger than 2^w) it's the same for sum - y.
2. two positive number and get a negtive number.
In this case : x + y > 2^(w-1) - 1 => since the maximum number of int is 2^(w-1) - 1
so sum = (x + y - 2^w => for example 5 + 5 will get -6 if w = 4
so when you use sum - x you will get (y - 2^w) since y < 2^(w-1) - 1 so the value will be
smaller than -1 - 2(w-1) which is also over flow so in fact it is (y - 2^w) % 2^w == y.
*/
难道不是永远返回非0值吗
整形加法本来就有可能溢出啊