机器码真数原码反码补码 float的内存表示

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#include<stdio.h>int main(int argc, char** argv){    printf("sizeof(int): %d\n", sizeof(int));    /*********    hex格式表示的整数是补码    http://www.cnblogs.com/zhangziqiu/archive/2011/03/30/ComputerCode.html    为简洁，计算机没有减法、只有加法; 反码解决减法问题；补码在反码基础上解决0的符号问题(然后用-0表示-128)    正数的反码是其本身      负数的反码是在其原码的基础上, 符号位不变，其余各个位取反.    正数的补码就是其本身    负数的补码是在其原码的基础上, 符号位不变, 其余各位取反, 最后+1. (即在反码的基础上+1)    [+1] = [00000001]原 = [00000001]反 = [00000001]补    [-1] = [10000001]原 = [11111110]反 = [11111111]补    因为机器使用补码, 所以对于编程中常用到的32位int类型, 可以表示范围是: [-231, 231-1] 因为第一位表示的是符号位.而使用补码表示时又可以多保存一个最小值    ********/    printf("0xFFFFFFFF:\t%d 0x%X\n", 0xFFFFFFFF, 0xFFFFFFFF);    printf("0x7FFFFFFF:\t%d 0x%X\n", 0x7FFFFFFF, 0x7FFFFFFF);    printf("-0x7FFFFFFF:\t%d 0x%X\n", -0x7FFFFFFF, -0x7FFFFFFF);        return 0;}

[root@test-22 leetcode]# !g++g++ -o inttest inttest.cpp ; ./inttest sizeof(int): 40xFFFFFFFF:     -1 0xFFFFFFFF0x7FFFFFFF:     2147483647 0x7FFFFFFF-0x7FFFFFFF:    -2147483647 0x80000001

https://github.com/xiarendeniao/utils/blob/master/c/verfloat.cpp

#include <stdio.h>#include <stdint.h>#include <string.h>#include <assert.h>typedef int64_t int64;typedef int32_t int32;typedef int16_t int16;typedef int8_t int8;typedef uint64_t uint64;typedef uint32_t uint32;typedef uint16_t uint16;typedef uint8_t uint8;typedef uint32_t DWORD;/*1.int32表示范围和机器码(补码) bin()函数用于验证补码表示形式    int32 [-2^31,2^31-1] [-2147483648,2147483647]    -2147483648正数部分超出31bit的表示范围,机器表示形式是-0(10000000 00000000 00000000 00000000)    所以不能用“符号位不变、其余按照正数的二进制形式按位取反再加1”获取-2147483648的补码 2.大头小头(大端序/小端序) big endian / little endian (以下以int32举例)    机器数(补码)只是为了方便运算所约定的计算机底层的数字表示形式，不等同于内存布局,内存布局和大头小头相关    dec(10)   = bin(0b00000000 00000000 00000000 00001010)    = hex(0xA)            hex是补码的十六进制    dec(-10)  = bin(0b11111111 11111111 11111111 11111011)    = hex(0xA)    dec(-1)   = bin(0x11111111 11111111 11111111 11111111)    = hex(0xFFFFFFFF)        在big endian机器的内存布局(数字的高位部分放在低地址处、低位部分放在高地址处) membin()函数用于验证数字在内存中的形式    dec(10)     00000000 00000000 00000000 00001010 即 00 00 00 0A    dec(-10)    11111111 11111111 11111111 11111011 即 FF FF FF FB    dec(-1)     11111111 11111111 11111111 11111111 即 FF FF FF FF    在little endian机器的内存布局(数字的高位部分放在高地址处、低位部分放在低地址处)    dec(10)     00001010 00000000 00000000 00000000 即 0A 00 00 00    dec(-10)    11111011 11111111 11111111 11111011 即 FB FF FF FF    dec(-1)     11111111 11111111 11111111 11111111 即 FF FF FF FF*/char* positive2bin(const int in, char* binstr, bool filling = true){      assert(in >= 0);    char* out = binstr;    char t[33];    char* tp = t;      int r = in;     while(r >= 1) {          *tp++ = r%2 + 48;          r /= 2;      }    *tp-- = '\0';    assert(tp-t == strlen(t)-1);    //fprintf(stderr, "Positive2bin:%12d 0b%s %dbits\n", in, t, strlen(t));     if (filling) for (int i = 0; i < 32-(tp-t+1); i++) *out++ = '0';    while (tp >= t) *out++ = *tp--; //BUG: while(*out++ = *tp--);    *out = '\0';        assert(out-binstr == strlen(binstr));    //fprintf(stderr, "--------2bin:%12d 0b%s %dbits\n", in, binstr, strlen(binstr));     return binstr;} char* bin(const int in, char* binstr){    if (in >= 0) return positive2bin(in, binstr);    positive2bin(-in, binstr);    //按位取反    *binstr = '1';    char* p = binstr + 1;    while ( *p != '\0') {        if (*p == '0') *p = '1';        else *p = '0';        p++;    }    p--;    //fprintf(stderr, "-----reverse:%12d 0b%s %dbits\n", in, binstr, strlen(binstr));    //+1    while (p > binstr) {        if (*p == '0') {            *p = '1';            break;        } else {            *p = '0';            p--;        }    }    if (p == binstr)         fprintf(stderr, "overflow: %d\n", in);    return binstr;}char* membin(const uint32* in, char* binstr){    char* bp = binstr;    char byte[9] = {0};    uint8* p = (uint8*)in;    for (int i = 0; i < 4; i++, p++) {        //printf("fetched byte %d\n", (uint32)*p);        positive2bin(*p, byte, false);        char* tp = byte;        for (int i = 0; i < 8-strlen(byte); i++) *bp++ = '0';        while (*tp != '\0') *bp++ = *tp++;    }    *bp = '\0';    //fprintf(stderr, "------v2-bin:%12d 0b%s %dbits\n", in, binstr, strlen(binstr));    return binstr;}//验证补码的表示形式跟内存数据存放(big/little endian)void verf_bin(){    char buff[128] = {0};    char binstr[33] = {0};    uint32 in;begin:    printf("please input > ");      scanf("%s", buff);    if (buff[0] == '0' && (buff[1] == 'x' || buff[1] == 'X'))        sscanf(buff, "%x", &in);    else        sscanf(buff, "%d", &in);    if (in != 0x80000000U) //-2147483648，int32的最小值        fprintf(stderr, "positive2bin:%12d 0x%08x 0b%s bin %f\n", in, in, bin((int32)in, binstr), *(float*)&in);    fprintf(stderr, "memory--2bin:%12d 0x%08x 0b%s mem %f\n", in, in, membin((uint32*)&in, binstr), *(float*)&in);    goto begin;  }int main(int argc, char** argv){    char binstr[33] = {0};    uint32 a = 0x7f700001U;    printf("0x%8x 0b%s %f\n", a, bin((int)a,binstr), *(float*)&a);    //https://android.googlesource.com/platform/dalvik.git/+/f6c387128427e121477c1b32ad35cdcaa5101ba3/libcore/luni/src/main/native/java_lang_Float.c    //java源码中把所有nan替换成同一个，这样便于做比较    uint32 nan = 0x7fc00000;    printf("+float标准的nan 0x%8x 0b%s %f\n", nan, bin((int)nan,binstr), *(float*)&nan);    uint32 b = 0x7f800001U;    printf("+float最小的nan 0x%8x 0b%s %f\n", b, bin((int)b,binstr), *(float*)&b);    uint32 c = 0x7fffffffU;    printf("+float最大的nan 0x%8x 0b%s %f\n", c, bin((int)c,binstr), *(float*)&c);    uint32 d = 0xff800001U;    printf("-float最大的nan 0x%8x 0b%s %f\n", d, bin((int)d,binstr), *(float*)&d);    uint32 e = 0xffffffffU;    printf("-float最小的nan 0x%8x 0b%s %f\n", e, bin((int)e,binstr), *(float*)&e);    verf_bin();    return 0;}/*[root@test-22 c++-study]# g++ verfloat.cpp -o verfloat; ./verfloat 0x7f700001 0b01111111011100000000000000000001 319014739270789413148584118414533984256.000000+float最小的nan 0x7f800001 0b01111111100000000000000000000001 nan+float最大的nan 0x7fffffff 0b01111111111111111111111111111111 nan-float最大的nan 0xff800001 0b11111111100000000000000000000001 nan-float最小的nan 0xffffffff 0b11111111111111111111111111111111 nanplease input > 0x40000001Upositive2bin:  1073741825 0x40000001 0b01000000000000000000000000000001 bin 2.000000memory--2bin:  1073741825 0x40000001 0b00000001000000000000000001000000 mem 2.000000please input > 0x7f000001Upositive2bin:  2130706433 0x7f000001 0b01111111000000000000000000000001 bin 170141203742878835383357727663135391744.000000memory--2bin:  2130706433 0x7f000001 0b00000001000000000000000001111111 mem 170141203742878835383357727663135391744.000000please input > 0x00000001Upositive2bin:           1 0x00000001 0b00000000000000000000000000000001 bin 0.000000memory--2bin:           1 0x00000001 0b00000001000000000000000000000000 mem 0.000000please input > 0x007fffffUpositive2bin:     8388607 0x007fffff 0b00000000011111111111111111111111 bin 0.000000memory--2bin:     8388607 0x007fffff 0b11111111111111110111111100000000 mem 0.000000please input > 0x3f800000Upositive2bin:  1065353216 0x3f800000 0b00111111100000000000000000000000 bin 1.000000memory--2bin:  1065353216 0x3f800000 0b00000000000000001000000000111111 mem 1.000000*//*    1.NaN        https://android.googlesource.com/platform/dalvik.git/+/f6c387128427e121477c1b32ad35cdcaa5101ba3/libcore/luni/src/main/native/java_lang_Float.c        上述URL中NAN的判定规则：指数为无效值            1.xxxx * 2^n            符号    指数(n,+/-)                         尾数(1.xxxx..)    float(32)               1       8                                   23                                    ...    0x7f800001U             0       111 1111 1                          000 0000 0000 0000 0000 0001    nan    0x7fffffffU             0       111 1111 1                          111 1111 1111 1111 1111 1111    nan    0xff800001U             1       111 1111 1                          000 0000 0000 0000 0000 0001    nan    0xffffffffU             1       111 1111 1                          111 1111 1111 1111 1111 1111    nan    0x7fc00000U             0       111 1111 1                          100 0000 0000 0000 0000 0000    nan        0x40000001U             0       100 0000 0                          000 0000 0000 0000 0000 0001        0x7f000001U             0       111 1111 0                          000 0000 0000 0000 0000 0001        0x00000001U             0       000 0000 0                          000 0000 0000 0000 0000 0001    -> 指数000 0000 0表示的是-127    0x007fffffU             0       000 0000 0                          111 1111 1111 1111 1111 1111        0x3f800000U             0       011 1111 1                          000 0000 0000 0000 0000 0000    -> 指数011 1111 1表示的是0，尾数全零表示的是1(尾数最高位被约定省略了)                                        [-2^7,2^7-1]                                    [-128,127](转正：约定+127后存入指数区间，即，127左移一位)                                    [-1,254]                                    [?,1111 1110] -1怎么存？ -> 应该是舍掉了吧，指数为0(即1.xxx * 2^-127)时浮点的值已经趋近于0了                                        double(64)              1       11                                  52                                    [-2^10,2^10-1]                                    [-1024,1023]                                    [-1,2046](转正：约定+1023后存入指数区间，即，1023左移一位)                                        0x7ff8000000000000ULL   0       111 1111 1111                       1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000*/

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机器码 真数 原码 反码 补码 float的内存表示

机器码真数原码反码补码 float的内存表示