在 C 中将字符 * 转换为无符号字符数组

Question

我想优化我的代码，避免错误，我有这个功能，但我认为我可以改善和避免内存问题。

void function(char* message)
{
    char * pointer;
    unsigned char buffer[2048] = {0};
    int buffer_len = 0;

    memset(buffer, 0, sizeof(buffer));
    strcpy(buffer, message);

    buffer_len = strlen(buffer);
    memset(buffer, 0, sizeof(&buffer));

    for(int i = 0, pointer = message; i < (buffer_len / 2); i++, pointer += 2)
    {
        sscanf(pointer, "%02hhX", &buffer[i]);
    }
}

该函数的思想是接收这种样式的字符串“0123456789”并将其传递给0x01、0x23、0x45、...在无符号字符数组中。任何提示，良好的实践或改进将是非常有用的。

用法是这样的：

function("0123456789");

在函数缓冲区中结束如下：

buffer[0] = 0x01
buffer[1] = 0x23
...

Answer 1

char * a= -80; // let supposed you get returned value( i represented in int) from whatever source.
unsigned char b = * a; // now b equal the complemntry of -80 which will be 176
std::cout << b ;

Answer 2

有一些可能的优化。最大的优化来自避免做2x内存集和strcpy，

无需：

// memset(buffer, 0, sizeof(buffer)); 
// strcpy(buffer, message);
// memset(buffer, 0, sizeof(&buffer));  

这大大简化了代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void function(char* message)
{
    unsigned char * pointer;
    unsigned char buffer[2048]; // not needed = {0}; 
    int buffer_half_len; // not needed = 0;

    buffer_half_len = strlen(message)/2;  // avoiding division in the for loop;
    pointer = message;

    for(int i = 0;  i < buffer_half_len; i++, pointer += 2)
    {
        sscanf(pointer, "%02hhX", &buffer[i]);
        printf("%02hhX\n", buffer[i] );
    }
}

输出:

01
23
45
67
89