将未签名的char *（uint8_t *）转换为const char * [英] cast unsigned char * (uint8_t *) to const char *

问题描述

我有一个使用uint8_t *参数的函数：

I've a function which take an uint8_t * argument :

uint8_t* ihex_decode(uint8_t *in, size_t len, uint8_t *out)
{
    uint8_t i, hn, ln;

    for (i = 0; i < len; i+=2) {
        hn = in[i] > '9' ? (in[i]|32) - 'a' + 10 : in[i] - '0';
        ln = in[i+1] > '9' ? (in[i+1]|32) - 'a' + 10 : in[i+1] - '0';

        out[i/2] = (hn << 4 ) | ln;
    }

    return out;
}

我将此功能与：

uint8_t data[SPM_PAGESIZE]; // SPM_PAGESIZE = 256 bytes
uint8_t sysex_data[SPM_PAGESIZE/2];
ihex_decode(data, strlen(data), sysex_data);

但是在这种情况下，我的编译器（avr-gcc）返回警告：

But in this case, my compiler (avr-gcc) return a warning :

main.c | 89 |警告：传递 strlen的参数1的指针目标的签名不同
/usr/include/string.h | 399 |注：预期为'const char *'，但参数的类型为'uint8_t *'

main.c|89|warning: pointer targets in passing argument 1 of 'strlen' differ in signedness /usr/include/string.h|399|note: expected 'const char *' but argument is of type 'uint8_t *'

因此，我发现了通过类型转换数据var的解决方案：

So, i've found a solution by type casting the data var :

ihex_decode(data, strlen((const char *)data), sysex_data);

警告消失了，但我想知道这种解决方案是否安全。

The warning disappears but I wonder if this solution is safe.

有更好的方法吗？

谢谢

推荐答案

这很安全。该错误与将8位无符号整数与字符混合在一起，如果只使用 char ，则这些字符是有符号的。

It is safe. The error has to do with mixing unsigned integers of 8 bits with characters, that are signed if you use just char.

但是，我看到该函数接受 uint8_t 并执行 char acter算术运算，因此它应该接受 char s（或 const char s）。请注意，字符常量'c'的类型为 char ，并且您在表达式中混合了有符号和无符号在 ihex_decode 内，因此您必须小心避免溢出或负数被视为大正数。

I see, however, that the function accepts uint8_t and does character arithmetic, so it should accepts chars (or const chars, for the matter). Note that a character constant 'c' is of type char, and you're mixing signed and unsigned in the expressions inside ihex_decode, so you have to be careful to avoid overflows or negative numbers treated as big positive numbers.

A最后的样式说明。由于未修改，因此应在中读取 const uint8_t *（或 const char *在（如上）中。另一个样式错误（可能会导致非常严重的错误）是您接受 len 作为 size_t ，但是声明 i 循环变量为 uint8_t 。如果该字符串的长度超过255个字节怎么办？

A last style note. As in is not modified, it should read const uint8_t* in (or const char* in, as of above) in the parameters. Another style error (that may lead to very bad errors) is that you accept len as size_t, but declare the i loop variable as uint8_t. What if the string is more than 255 bytes in length?

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