把字符数组转换成字符串

[atpeak]

假设BYTE BUFFER[20]里存放了20个字节的数据。
比如BUFFER里的数据可能是0x3B 0x6F 0x00 0x00 0x56 ……
现在需要把BUFFER里的数据以3B6F000056…的形式打印在屏幕上。
请问怎么用C实现？
谢谢您了！

[baihacker]

#include <iostream>
#include <iomanip>
using namespace std;

int main()
{
	unsigned char buff[] = {0x3B, 0x6F, 0x00, 0x00, 0x56};
	for (int i = 0; i < sizeof(buff)/sizeof(unsigned char); ++i)
		cout << "0x" <<hex << setfill('0') << setw(2)
			<< (buff[i] & 0xff) << '\t';
	return 0;
}

[baihacker]

#include <stdio.h>

int main()
{
	unsigned char buff[] = {0x3B, 0x6F, 0x00, 0x00, 0x56};
	int	i;
	for (i = 0; i < sizeof(buff)/sizeof(unsigned char); ++i)
		printf("0X%02X\t", buff[i]);
	return 0;
}

[atpeak]

谢谢baihacker飞雪！

现在把问题深入一步，如果要把BUFFER里的数据
unsigned char buff1[] = {0x3B, 0x6F, 0x00, 0x00, 0x56};
转换成
unsigned char buff2[] = {'3','B','6','F','0','0',’0','0','5','6'};
应该如何用C实现呢？

[jieao111]

飞雪的不符合要求吧，，你不是把数组理的0x，都要去掉吗

[baihacker]

#include <stdio.h>

int main()
{
    unsigned char buff[] = {0x3B, 0x6F, 0x00, 0x00, 0x56};
	char	hex_tb[]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
	char		result[256];
    int			i;
	char		t;
    for (i = 0; i < sizeof(buff)/sizeof(unsigned char); ++i)
        result[i*2] = hex_tb[(buff[i]>>4)&0x0f],
		result[i*2+1] = hex_tb[buff[i]&0xf];
	for (i = 0; i < sizeof(buff)/sizeof(unsigned char)*2; ++i)
		printf("%c ", result[i]);
    return 0;
}

[baihacker]

//刚才看错要求了
#include <stdio.h>

int main()
{
    unsigned char buff[] = {0x3B, 0x6F, 0x00, 0x00, 0x56};
    int    i;
    for (i = 0; i < sizeof(buff)/sizeof(unsigned char); ++i)
        printf("%02X", buff[i]);
    return 0;
}

[baihacker]

//C++的是这样
#include <iostream>
#include <iomanip>
using namespace std;

int main()
{
    unsigned char buff[] = {0x3B, 0x6F, 0x00, 0x00, 0x56};
    for (int i = 0; i < sizeof(buff)/sizeof(unsigned char); ++i)
        cout  << setfill('0') << setw(2) << hex << uppercase
            << (buff[i] & 0xFF);
    return 0;
}