1. 如何自己计算得到边界值

注：整数常量的数据类型默认为：能表示该数的最小数据类型。

以下从小到大排列，选择表中能够表示该整数常量的第一个数据类型（如 8 选 int，2^32+10 选 long int）

- 10进制常量可选：int, long int, long long int

- 8进制或16进制常量可选：int, unsigned int，long int,unsigned long int, long long int,unsigned long long int）

// 计算数据类型的边界：最小值和最大值
//
#include <iostream>
using namespace std;
int main()
{
    //计算无符号整型
    unsigned int uint_max, uint_min;
    uint_max = ~(unsigned)0;//按位取反
    uint_min = 0;
    cout <<"uint_max = " << uint_max << endl;
    cout <<"uint_min = " << uint_min << endl;
    //计算有符号整型
    int int_max, int_min;
    int_max = (~(unsigned)0 >> 1) ;
    int_min = -int_max - 1;
    cout <<"int_max = "<< int_max << endl;
    cout <<"int_min = "<< int_min << endl;
}

2. C/C++库文件的数据类型边界

在C语言中的各数据类型的边界如下：

包含在头文件#include <limits.h>

c++包含在#include <limits> – 引用的是limits.h的内容

//limits.h
#define MB_LEN_MAX    5             /* max. # bytes in multibyte char */
#define SHRT_MIN    (-32768)        /* minimum (signed) short value */
#define SHRT_MAX      32767         /* maximum (signed) short value */
#define USHRT_MAX     0xffff        /* maximum unsigned short value */
#define INT_MIN     (-2147483647 - 1) /* minimum (signed) int value */
#define INT_MAX       2147483647    /* maximum (signed) int value */
#define UINT_MAX      0xffffffff    /* maximum unsigned int value */
#define LONG_MIN    (-2147483647L - 1) /* minimum (signed) long value */
#define LONG_MAX      2147483647L   /* maximum (signed) long value */
#define ULONG_MAX     0xffffffffUL  /* maximum unsigned long value */
#define LLONG_MAX     9223372036854775807i64       /* maximum signed long long int value */
#define LLONG_MIN   (-9223372036854775807i64 - 1)  /* minimum signed long long int value */
#define ULLONG_MAX    0xffffffffffffffffui64       /* maximum unsigned long long int value */

3. 如何判读溢出

参考传送门：

C语言的整型溢出问题：http://www.kuqin.com/shuoit/20140421/339376.html

#include <limits.h>
//有符号整型的判断溢出
void f(signed int si_a, signed int si_b) {
    signed int sum;

    /* Handle error */
    if (((si_b > 0) && (si_a > (INT_MAX - si_b))) ||
        ((si_b < 0) && (si_a < (INT_MIN - si_b)))) {

        return;
    }

    sum = si_a + si_b;
}