如何在C中将二进制编码的十进制转换为int++

一个简单的实现如下:

int from_hex(uint16_t h)
{
    int d = 0;
    int power = 1;
    while (h)
    {
        // assert(h % 16 < 10)
        d += h % 16 * power;
        h /= 16;
        power *= 10;
    }
    return d;
}

多亏了 Minor Threat 对于以下各项:

int from_hex(uint16_t h)
{
    int d = 0;
    int power = 1;
    while (h)
    {
        // assert(h % 16 < 10)
        d += (h & 15) * power;
        h >>= 4;
        power *= 10;
    }
    return d;
}

我很高兴知道我的代码是否有问题

既然你说十六进制数从不包含“a”、“b”等,那么这段代码应该做到这一点:

#include <iostream>

int main() {
    uint16_t in = 0x4321;

    int t = (1000 * ((in & 0xf000) / (16*16*16))) + 
             (100 * ((in & 0xf00) / (16*16))) + 
              (10 * ((in & 0xf0) / 16)) + 
               (1 * ((in & 0xf) / 1));
    std::cout << t << std::endl;
    return 0;
}

4321

解释

=0xWZYX中的16位十六进制数计算如下:

in = W*16^3 + Z*16^2 + Y*16^1 + X*16^0 (or just W*16^3 + Z*16^2 + Y*16^1 + X)

in & 0xf000 you get 0xW000

执行时

0xW000 / 16^3 you get 0x000W or just W

执行时

1000 * W you get W000 decimal

然后对每个数字重复该模式。

使用shift的替代实现

#include <iostream>

int main() {
    uint16_t in = 0x9321;

    int t = (1000 * ((in & 0xf000) >> 12)) + 
             (100 * ((in & 0xf00) >> 8)) + 
              (10 * ((in & 0xf0) >> 4)) + 
               (1 * ((in & 0xf) >> 0));
    std::cout << t << std::endl;
    return 0;
}

对于16位无符号整数,可以写出四行。然而,如果您希望为较大的无符号int使用类似的函数,那么最好执行循环以使代码更加紧凑和可维护。

#include <iostream>

int64_t directHexToDec(uint64_t in)
{
    int64_t res = 0;
    uint64_t mask = 0xf;
    uint64_t sh = 0;
    uint64_t mul = 1;
    for (int i=0; i<16; ++i)
    {
        res += mul * ((in & mask) >> sh);
        mul *= 10;
        mask <<= 4;
        sh += 4;
    }
    return res;
}

int main() {
    uint64_t in = 0x987654321;

    int64_t t = directHexToDec(in);
    std::cout << t << std::endl;
    return 0;
}

输出

987654321

0x0A 0X0B , ... 价值观

uint16_t val = 0x80;
std::stringstream stream;
stream << std::hex << val;
std::string resultStr(stream.str());
// In case of 0xYY hex value be carefull that YY fits into an uint8_t, 
// otherwise  this will overflow
uint8_t result = static_cast<uint8_t>(std::stoi(resultStr));

这应该适用于十六进制<=0x99:

#include <iostream>

unsigned int foo(unsigned int hex)
{
    return  hex - (hex >> 4) * 6;
}

/* Test */
int main()
{
    unsigned int vals[] = {0x08, 0x09, 0x10, 0x11, 0x12, 0x80};

    for (int i = 0; i < sizeof(vals) / sizeof(vals[0]); i++)
        std::cout << foo(vals[i]) << std::endl;

    return 0;
}

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80