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当int类型超出了[-2147483648,2147483647]会发生什么?

当int超超出范围,编译器会报错吧?。。。。

答案是不会(看编译器怎么处理),在vs环境中,当输入下面的代码:

        int a = 2147483647;
	int b = 2147483648;
	int d = 2147483649;
	int e = 2147483650;

结果是:

也就是说2147483648变成了-2147483648,即循环赋值了,一旦超出就循环到最小的值,依此类推。

那么在计算中怎么判断溢出呢,比如,有个循环不停累加,当溢出就返回一个结果。

我们用一个更大的数,比如__int64(两个32位数再怎么加、乘运算也不会溢出64位,比如4位的1111b乘以1111b(最高位为符号位)就是十进制的7*7,最大的也8位在-128~127之间)

有人说用long,long这个慎用,因为在32位机器和64位机器上表示范围不一样,具体可以通过下面的方法查询:

        printf("int取值范围:%d ~ %d\n", INT_MIN, INT_MAX);
	printf("long取值范围:%d ~ %d\n", LONG_MIN, LONG_MAX);

头文件是:

#include <limits.h>

或者直接打开这个头文件:(我这是32位的机器)

//
// limits.h
//
//      Copyright (c) Microsoft Corporation. All rights reserved.
//
// The C Standard Library <limits.h> header.
//
#pragma once
#define _INC_LIMITS

#include <vcruntime.h>

_CRT_BEGIN_C_HEADER



#define CHAR_BIT      8         // number of bits in a char
#define SCHAR_MIN   (-128)      // minimum signed char value
#define SCHAR_MAX     127       // maximum signed char value
#define UCHAR_MAX     0xff      // maximum unsigned char value

#ifndef _CHAR_UNSIGNED
    #define CHAR_MIN    SCHAR_MIN   // mimimum char value
    #define CHAR_MAX    SCHAR_MAX   // maximum char value
#else
    #define CHAR_MIN    0
    #define CHAR_MAX    UCHAR_MAX
#endif

#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

#define _I8_MIN     (-127i8 - 1)    // minimum signed 8 bit value
#define _I8_MAX       127i8         // maximum signed 8 bit value
#define _UI8_MAX      0xffui8       // maximum unsigned 8 bit value

#define _I16_MIN    (-32767i16 - 1) // minimum signed 16 bit value
#define _I16_MAX      32767i16      // maximum signed 16 bit value
#define _UI16_MAX     0xffffui16    // maximum unsigned 16 bit value

#define _I32_MIN    (-2147483647i32 - 1) // minimum signed 32 bit value
#define _I32_MAX      2147483647i32 // maximum signed 32 bit value
#define _UI32_MAX     0xffffffffui32 // maximum unsigned 32 bit value

// minimum signed 64 bit value
#define _I64_MIN    (-9223372036854775807i64 - 1)
// maximum signed 64 bit value
#define _I64_MAX      9223372036854775807i64
// maximum unsigned 64 bit value
#define _UI64_MAX     0xffffffffffffffffui64

#if _INTEGRAL_MAX_BITS >= 128
    // minimum signed 128 bit value
    #define _I128_MIN   (-170141183460469231731687303715884105727i128 - 1)
    // maximum signed 128 bit value
    #define _I128_MAX     170141183460469231731687303715884105727i128
    // maximum unsigned 128 bit value
    #define _UI128_MAX    0xffffffffffffffffffffffffffffffffui128
#endif

#ifndef SIZE_MAX
    #ifdef _WIN64
        #define SIZE_MAX _UI64_MAX
    #else
        #define SIZE_MAX UINT_MAX
    #endif
#endif

#if __STDC_WANT_SECURE_LIB__
    #ifndef RSIZE_MAX
        #define RSIZE_MAX (SIZE_MAX >> 1)
    #endif
#endif



_CRT_END_C_HEADER

可以看出 long long和__int64是一样的。

说到这,还有个问题:下面的a等于多少?

long long a = 2147483647 +1;

你可能觉得是2147483648,但结果却是-2147483648,why?。。。

因为2147483647是在int范围里,编译器把它当作int来处理,加1后结果还是int,超范围了。

这样结果就是:2147483648

long long a = (long long)2147483647 +(long long)1;