当int类型超出了[ |
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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 header. // #pragma once #define _INC_LIMITS #include _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;
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