c++ - 编辑存储在字节数组中的数据

Question

我一直在尝试将经常操作的数据存储到一个大的无符号字符数组中(因为 C++ 没有字节类型)。所以，如果我将一个 float 存储到 char 数组中，它将占用 4 个无符号字符。很简单，除了现在如果我想编辑数组中的数据，我需要同时访问所有 4 个元素，据我所知这是不可能的。那么，有没有一种方法可以在不求助于 memcpy() 的情况下将 4 个无符号字符编辑为我想要的浮点值？

例子:

#include <iostream>
#include <string.h>
#include <stdint.h>

using namespace std;

struct testEntity {

    float density;
    uint16_t xLoc, yLoc;
    uint16_t xVel, yVel;
    uint16_t xForce, yForce;
    uint16_t mass;
    uint8_t UId;
    uint8_t damage; 
    uint8_t damageMultiplier;
    uint8_t health;
    uint8_t damageTaken;
};


int main()
{

    testEntity goblin { 1.0, 4, 5, 5, 0, 0, 0, 10, 1, 2, 1, 10, 0 };
    testEntity goblin2;
    unsigned char datastream[24];
    unsigned char* dataPointer = datastream;

    memcpy(&datastream, &goblin, sizeof(testEntity));


    //I know that datastream[0..3] contains information on goblin.density
    //How would I edit goblin.density without memcpy into another testEntity structure?

    memcpy(&goblin2, &datastream, sizeof(testEntity));

return 0;
}

Answer 1

这是我做的:

#include <iostream>
#include <string.h>
#include <stdint.h>

using namespace std;

struct testEntity {

    float density;
    uint16_t xLoc, yLoc;
    uint16_t xVel, yVel;
    uint16_t xForce, yForce;
    uint16_t mass;
    uint8_t UId;
    uint8_t damage; 
    uint8_t damageMultiplier;
    uint8_t health;
    uint8_t damageTaken;
};


int main()
{

    testEntity goblin = { 1.0, 4, 5, 5, 0, 0, 0, 10, 1, 2, 1, 10, 0 };
    testEntity goblin2;
    unsigned char datastream[24];
    unsigned char* dataPointer = datastream;
    testEntity *goblinP;

    memcpy(datastream, &goblin, sizeof(testEntity));


    goblinP = (testEntity *) datastream;

    cout << goblinP->density << endl;

return 0;
}