将char数组转换为十六进制数组

Question

输入：

char arr1[9] = "+100-200"  // (+ is 2B, - is 2D, 1 is 31 and 2 is 32)

输出：

unsigned int arr2[4]= [0x2B31,0x3030,0x2D32,0x3030]

我该怎么做？

Answer 1

您的问题似乎不一致：0应转换为0x30，即其ASCII值。

为什么要进行这种修改，代码非常简单：

char arr1[8] = "+100-200";
unsigned int arr2[4];

for (int i = 0; i < 8; i += 2) {
    arr2[i / 2] = ((unsigned int)(unsigned char)arr1[i] << 8) |
                   (unsigned int)(unsigned char)arr1[i + 1];
}

for (int i = 0; i < 4; i++) {
    printf("0x%04X ", arr2[i]);
}
printf("\n");

输出：

0x2B31 0x3030 0x2D32 0x3030