使用malloc（）为const char字符串动态分配内存， [英] Dynamically allocating memory for const char string using malloc()

问题描述

我编写一个程序，从.ini文件读取一个值，然后将值传递给接受PCSTR（即const char *）的函数。函数 getaddrinfo（）。

因此，我想写 PCSTR ReadFromIni ）。要返回一个常量字符串，我计划使用 malloc（）分配内存并将内存转换为常量字符串。我将能够获得从.ini文件读取的字符的确切数量。

这种技术好吗？

以下示例在Visual Studio 2013中运行良好，并根据需要打印hello。

  const char * m（）
 {
 char * c =（char *）malloc * sizeof（char））; 
 c =hello; 
 return（const char *）c; 
} 
 
 int main（int argc，char * argv []）
 {
 const char * d = m（）; 
 std :: cout<< d; // use PCSTR 
} 
  

解决方案

第二行是可怕的错误：

  char * c =（char *）malloc sizeof（char））; 
 //'c'设置为指向一块分配的内存（通常位于堆中）
 c =hello; 
 //'c'设置为指向一个常量字符串（通常位于代码段或数据部分）
  

你分配变量 c 两次，所以很明显，第一个赋值没有意义。
它就像写：

  int i = 5; 
 i = 6; 
  

最重要的是，你会丢失分配的内存的地址，所以你不会

$ b

p $ p> char * m（）
{
const char * s =hello;
char * c =（char *）malloc（strlen（s）+1）;
strcpy（c，s）;
return c;请记住，任何调用 char * p =





$ b < m（）也必须在以后调用 free（p） ...

I am writing a program that reads a value from an .ini file, then passes the value into a function that accepts a PCSTR (i.e. const char *). The function is getaddrinfo().

So, I want to write PCSTR ReadFromIni(). To return a constant string, I plan on allocating memory using malloc() and casting the memory to a constant string. I will be able to get the exact number of characters that were read from the .ini file.

Is that technique okay? I don't really know what else to do.

The following example runs fine in Visual Studio 2013, and prints out "hello" as desired.

const char * m()
{
    char * c = (char *)malloc(6 * sizeof(char));
    c = "hello";
    return (const char *)c;
}    

int main(int argc, char * argv[])
{
    const char * d = m();
    std::cout << d; // use PCSTR
}

解决方案

The second line is "horribly" wrong:

char* c = (char*)malloc(6*sizeof(char));
// 'c' is set to point to a piece of allocated memory (typically located in the heap)
c = "hello";
// 'c' is set to point to a constant string (typically located in the code-section or in the data-section)

You are assigning variable c twice, so obviously, the first assignment has no meaning. It's like writing:

int i = 5;
i = 6;

On top of that, you "lose" the address of the allocated memory, so you will not be able to release it later.

You can change this function as follows:

char* m()
{
    const char* s = "hello";
    char* c = (char*)malloc(strlen(s)+1);
    strcpy(c,s);
    return c;
}

Keep in mind that whoever calls char* p = m(), will also have to call free(p) at some later point...

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