c-在函数中初始化的数组的指针在返回后包含错误的值

Question

I'm facing an issue and I'm certainly doing something wrong. I need to call a function that returns a pointer to an array of int but when after it returns, the values inside the array are wrong and some values are missing.

int* patternForFirstDigit(int digit) {
    int *pattern;
    pattern = (int [6]){1,1,1,1,1,1};

    switch (digit) {
        case 0:
            pattern = (int [6]){1,1,1,1,1,1};
            break;

        case 1:
            pattern = (int [6]){1,1,2,1,2,2};
            break;

        default:
            pattern = (int [6]){0,0,0,0,0,0};
            break;
    }

    for (int i = 0; i < 6; i++) {
         printf("%i\n", pattern[i]);
    }

    return pattern;
}

In case of digit = 1, here's what's printed

1, 1, 2, 1, 2, 2

But after returning

int *pattern = patternForFirstDigit(0);
for (int i = 0; i < 6; i++) {
     printf("%i\n", pattern[i]);
}

here's what's printed

1, -1405451528, -1405449120, 366001

Do you have an idea of what's wrong ?

Thanks guys

PS : I'm using Xcode 4.6 and my project is using ARC but I'm pretty sure it's not the reason of my problem.

Answer 1

You can not return a pointer to an array created in a function. This array is no longer existing after a function returns, so your pointer points to some random, invalid place in a memory.

Allocate memory for a pointer (using malloc() for instance) and then return a pointer. This would also mean you'd need to free a pointer after you're done with it (with free() ).

Pseudocode for this would be something like:

int* patternForFirstDigit(int digit) {
  int *pattern = (int*) malloc(sizeof(int)*N);
  pattern[0] = 0;
  pattern[1] = 1;
  ...

  // Alternatively just create a local array and use a for-loop
  // to copy the contents to the pattern array.

  return pattern;
}

int *p = patternForFirstDigit(M);
// use p
free(p);

Answer 2

the

(int [6]){1,1,2,1,2,2};

is a local array defined in the function . so the data of the array could be erased when the function finish the execution. so that's why you get garbage values in your printf

1) Use malloc to allocate array instead

int* patternForFirstDigit(int digit) {
    int *pattern = malloc(6*sizeof(int));

    memcpy(pattern, (int [6]){1,1,1,1,1,1}, 6*sizeof(int));

    switch (digit) {
        case 0:
            memcpy(pattern, (int [6]){1,1,1,1,1,1}, 6*sizeof(int));
            break;

        case 1:
            memcpy(pattern, (int [6]){1,1,2,1,2,2};, 6*sizeof(int));
            break;

        default:
            memcpy(pattern, (int [6]){0,0,0,0,0,0}, 6*sizeof(int));
            break;
    }

    for (int i = 0; i < 6; i++) {
         printf("%i\n", pattern[i]);
    }

    return pattern;
}

and some where in your code when the pattern became useless then free it with free(pattern);

2) Or use static in the definition of the array:

int* patternForFirstDigit(int digit) {
    int *pattern; int i;
    static int A[6]={1,1,1,1,1,1};
    static int B[6]={1,1,2,1,2,2};
    static int C[6]={0,0,0,0,0,0};
    pattern = A;

    switch (digit) {
        case 0:
            pattern = A;
            break;

        case 1:
            pattern = B;
            break;

        default:
            pattern = C;
            break;
    }

    for (i = 0; i < 6; i++) {
         printf("%i\n", pattern[i]);
    }

    return pattern;
}

Answer 3

The problem is because you are returning a local variable. A local variable is a temporary one that is no longer available after its scope is gone. You may try the following:

int* patternForFirstDigit(int digit, int* pattern) {

    int* pattern1 = (int [6]){1,1,1,1,1,1};
int i;
    switch (digit) {
        case 0:
            pattern1 = (int [6]){1,1,1,1,1,1};
            break;

        case 1:
            pattern1 = (int [6]){1,1,2,1,2,2};
            break;

        default:
            pattern1 = (int [6]){0,0,0,0,0,0};
            break;
    }
memcpy(pattern,pattern1,6*sizeof(int));
    for ( i = 0; i < 6; i++) {
         printf("%i\n", pattern[i]);
    }
    return pattern;
}

Then you may use it like:

pattern = malloc(6*sizeof(int));
pattern=patternForFirstDigit(1, pattern);
int i;
for ( i = 0; i < 6; i++) {
     printf("%i\n", pattern[i]);
}
free(pattern);