无符号字符 * 数据作为 c++ 或 c 中函数的返回类型

Question

我正在使用带有 CLR（通用语言运行时）的 Visual Studio 2010。我想将 unsigned char * 从 compress 函数返回到 main 函数，但是 unsigned char * 和 char * 都给出了错误

“错误 7 错误 C2440：'='：无法从 'char *' 转换为 'unsigned char *' F:\4-2\Thesis\PROJECT\Database Compression Main\Database Compression 2\db_comp_main.cpp 21 “

unsigned char* compressor(char *data)
{
    unsigned char *compressed_string;
    //With some process I had compressed string of data variable as unsigned char* in variable compressed_string successfully and also printed here. Now want to return it from here.
    return compressed_string;
}
int main()
{ 
  unsigned char *main_data; 
  main_data=compressor("Muhammad Ashikuzzaman.Student from Khulna University Of Engineering And Technology from Bangladesh");// When I click the error from error list the cursor is put here beside str by visual studio.
  printf("%s",main_data);
}

需要从压缩函数返回 unsigned char * 类型的数据。请帮忙。

Answer 1

更改任何数据类型（函数参数*data或局部变量*main_data）以匹配类型。尝试这个

unsigned char *main_data;

或者....

无论如何，简单的类型转换都可以使其正常工作：

unsigned char* compressor(char *data)
{
    unsigned char *compressed_string;
    compressed_string = (unsigned char*) data;
    return compressed_string;
}
int main()
{
  char *main_data;
  main_data=(char*) compressor("Muhammad Ashikuzzaman.Student from Khulna University Of Engineering And Technology from Bangladesh");
  printf("%s",main_data);
}

或者...

（用 改变一切char*）

有三种字符类型：（普通）char，signed char和unsigned char。Anychar通常是一个 8 位整数*（在 C 语言中没有专用的“字符类型”），从这个意义上说， asigned和unsigned char具有有用的含义（通常等同于uint8_tand int8_t）。当用作文本意义上的字符时，请使用 a char（也称为普通字符）。这通常是一个，signed char但可以由编译器以任何一种方式实现。所以我认为char*在这里使用是安全的。

工作示例（视为char *数据类型）：

char* compressor(char *data)
{
    char *compressed_string;
    compressed_string = "compressed";
    // compressed_string = data;
    return compressed_string;
}
int main()
{ 
  char *main_data; 
  main_data=compressor("Muhammad Ashikuzzaman.Student from Khulna University Of Engineering And Technology from Bangladesh");
  printf("%s",main_data);
}

此外，为简单起见，您可以使用指针到指针

void compressor(char **data) {
    *data = "compressed"; // assuming the compressed data will be smaller in length of main data, so buffer overflow won't occur  
}

int main() {
    char *main_data = "Muhammad Ashikuzzaman.Student from Khulna University Of Engineering And Technology from Bangladesh";
    compressor(&main_data);
    printf("%s", main_data);
    return 0;
}

编辑

当您询问通过引用传递的替代方法时，除了通过引用传递将数组作为函数参数传递之外，别无他法。因为当你写的时候，void func(char str[])你可能会认为你正在执行pass by value。但是 g++ 编译器会优化语句void func(char &str[0])以避免将整个数组作为函数参数进行处理。因此，在所有情况下，您实际上都是通过引用传递的。