如何在ANSI C程序中返回字符串数组?

How can I return an array of strings in an ANSI C program?

如何在ANSI C程序中返回字符串数组?

For example:

例如:

#include<stdio.h>

#define SIZE 10

char ** ReturnStringArray()
{
    //How to do this?
}

main()
{
    int i=0;

    //How to do here???

    char str ** = ReturnStringArray();

    for(i=0 ; i<SIZE ; i++)
    {
        printf("%s", str[i]);
    }
}

3 个解决方案

#1

You could do the following. Error checking omitted for brevity

您可以执行以下操作。为了简便起见，省略了错误检查

char** ReturnStringArray() {
  char** pArray = (char**)malloc(sizeof(char*)*SIZE);
  int i = 0;
  for ( i = 0; i < SIZE; i++ ) {
    pArray[i] = strdup("a string");
  }
  return pArray;
}

Note that you'd need to correspondingly free the returned memory.

注意，您需要相应地释放返回的内存。

Additionally in your printf call you'll likely want to include a \n in your string to ensure the buffer is flushed. Otherwise the strings will get queued and won't be immediately printed to the console.

另外，在printf调用中，您可能希望在字符串中包含一个\n，以确保刷新缓冲区。否则，字符串将排队，不会立即打印到控制台。

char** str = ReturnStringArray();
for(i=0 ; i<SIZE ; i++)
{
    printf("%s\n", str[i]);
}

#2

Do it this way

这样做

#include<stdio.h>

#define SIZE 10

char ** ReturnStringArray()
{
    //How to do this?
    char **strList = (char **)malloc(sizeof(char*) * SIZE);
    int i = 0;
    if (strList != NULL){
         for (i = 0; i < SIZE; i++){
             strList[i] = (char *)malloc(SIZE * sizeof(char) + 1);
             if (strList[i] != NULL){
                sprintf(strList[i], "Foo%d", i);
             }
         }
    }
    return strList;
}

main()
{
    int i=0;

    //How to do here???

    char **str = ReturnStringArray();

    for(i=0 ; i<SIZE ; i++)
    {
        printf("%s", str[i]);
    }
}

• Problem 1: Your double pointer declaration was incorrect
• 问题1:您的双指针声明是错误的。
• Problem 2: You need to know the size of the string for each pointer in the double-pointer..
• 问题2:你需要知道双指针中每个指针的字符串大小。
• Problem 3: The onus is placed on you to free the memory when done with it..
• 问题3:你有责任释放记忆

The code sample above assumes that the maximum size of the string will not exceed the value of SIZE, i.e. 10 bytes in length...

上面的代码示例假设字符串的最大大小不会超过大小的值，即10字节的长度……

Do not go beyond the boundary of the double pointer as it will crash

不要超过双指针的边界，因为它会崩溃

#3

pls dont typecast the return of malloc, you have not included <stdlib.h> and as someone pointed out above lack of prototype will result in int being casted to char **. Accidently your program may or may not work at all.

请不要打印malloc的返回，您没有包含 和上面有人指出缺少原型将导致int被铸造为char ** *。你的程序可能会也可能不会工作。 。h>