我如何 return 一个字符数组并在 main 中打印它？

Question

我构建了一个获取整数参数和 return 字符数组的函数。

例如，对于参数 13，函数应该 return "0013"；对于 3 的参数，函数应该 return "0003".

但我没有收到错误消息，我无法使用 printf(getInstructionIndex(13));

显示此值

这是我的代码：

/* get Instruction Index how "0001" or "0010" */
char * getInstructionIndex(int index){

    char number1;
    char number2;
    char number3;

    char *str = (char *) malloc(sizeof(char) * 4);

        if(index < 10){

            number1 = '0';
            number2 = '0';
            number3 = '0';

            str[0] = number1;
            str[1] = number2;
            str[2] = number3;
            str[3] = index;

            return str;
        }
        else{
            if(index < 100 && index >= 10){

                number1 = '0';
                number2 = '0';

                str[0] = number1;
                str[1] = number2;
                str[2] = index;

                return str;

            }
            else
            {
                if(index < 1000 && index >= 100){
                    number1 = '0';

                    str[0] = '0';
                    str[1] = index;

                    return str;
                }
                else
                {
                    str[0] = index;

                    return str;
                }
            }
        }

        free(str);

}

这是我的主要内容：

int main(int argc, char *argv[]){

    printf("started\n");


    printf(getInstructionIndex(13)); /* i must see 0013*/

    printf("stopped\n");

    return 0;
}

Answer 1

这段代码中确实存在内存泄漏，free 应该延迟到打印数组之后，例如：

char *str = getInstructionIndex(13)
printf("%s", str); /* i must see 0013*/
free(str);

如果您想打印函数返回的字符串，您应该使用：

printf("%s", str);

还有一件事，您在 getInstructionIndex 方法中构造的值需要由字符串终止符 [=14=] 终止（否则您将在输出中得到垃圾）。

Answer 2

问题

string str allocated only 4 chars, not enough for null terminator

null terminator not added

complex logic

memory leak

missing format specifier

casting malloc return

调整后的代码

#include <stdio.h>
#include <stdlib.h>

/* get Instruction Index how "0001" or "0010" */
char * getInstructionIndex(int index){
  /* overflow + underflow handling */
  if(index < 0 || index > 9999)return 0;

  /* allocate enough for the null terminator.. */
  char *str = malloc(sizeof(*str) * 5);
  /* error handling */
  if(!str)
  {
    return 0;
  }

  /* simplify logic, use a nice format specifier */
  sprintf(str, "%04d", index);
  return str;
}

int main(int argc, char *argv[])
{
  char *instruction_index;
  printf("started\n");
  instruction_index = getInstructionIndex(13);
  if(!instruction_index)
  {
    // error handling here..
    return 0;
  }
  /* add format string.. */
  printf("%s\n", instruction_index); /* i must see 0013*/
  /* release the memory */
  free(instruction_index);
  printf("stopped\n");
  return 0;
}

输出

$ gcc -g test.c -o test
$ valgrind ./test
==2713== Memcheck, a memory error detector
==2713== Copyright (C) 2002-2013, and GNU GPL'd, by Julian Seward et al.
==2713== Using Valgrind-3.10.0.SVN and LibVEX; rerun with -h for copyright info
==2713== Command: ./test
==2713== 
started
0013
stopped
==2713== 
==2713== HEAP SUMMARY:
==2713==     in use at exit: 0 bytes in 0 blocks
==2713==   total heap usage: 1 allocs, 1 frees, 5 bytes allocated
==2713== 
==2713== All heap blocks were freed -- no leaks are possible
==2713== 
==2713== For counts of detected and suppressed errors, rerun with: -v
==2713== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 0 from 0)

参考

printf
valgrind quickstart
casting malloc

我如何 return 一个字符数组并在 main 中打印它？

How can I return a char array and print it in main?

c

arrays

main

function

char