将 int 存储在 char* [C] [英] Store an int in a char* [C]

问题描述

int main()
{
   int n = 56;
   char buf[10]; //want char* buf instead
   sprintf(buf, "%i", n);
   printf("%s\n", buf);

   return 0;
}

这段代码有效，我的问题是如果我想要相同的行为，buf 是 char* 怎么办?

This piece of code works, my question is what if i want the same behaviour, with buf being a char* ?

推荐答案

buf 是一个静态分配在堆栈上的数组.要使其成为 char * 类型，您需要动态分配内存.我想你想要的是这样的:

buf is an array statically allocated on the stack. For it to be of type char * you need to allocate dynamically the memory. I guess what you want is something like this:

int main()
{
   int n = 56;
   char * buf = NULL;
   buf = malloc(10 * sizeof(char));
   sprintf(buf, "%i", n);
   printf("%s\n", buf);

   // don't forget to free the allocated memory
   free(buf);

   return 0;
}

正如Thomas Padron-McCarthy"在评论中所指出的，您不应该强制转换 malloc() 的返回值.您可以在此处找到更多信息.您也可以完全删除 sizeof(char)，因为它总是由 1 删除.

As pointed out by 'Thomas Padron-McCarthy' in the comments, you should not cast the return of malloc(). You will find more info here. You could also remove completely the sizeof(char) as it will always by 1.

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