浅谈溢出

#include <stdio.h>#include <stdlib.h>#include <math.h>#define MAX_INT ((unsigned)(-1) >> 1)#define MIN_INT (MAX_INT + 1)int main(){  int n=2147483647;  printf("%d",n+1);    return 0;}

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输出：-2147483648

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实际上：如果整型超过了范围，则实际输出的应该是该数无符号范围最大值MOD的结果

#include <stdio.h>#include <stdlib.h>#include <math.h>#define MAX_INT ((unsigned)(-1) >> 1)#define MIN_INT (MAX_INT + 1)int main(){  int n=2147483650;  printf("%d",n);return 0;}

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在codeblocks中 int占4个byte
范围是0~4294967296

故结果应为2147483650-4294967296=-2147483646

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-2147483646Process returned 0 (0x0)   execution time : 0.036 sPress any key to continue.

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