What am I doing wrong with pointers in C?

RDP1998 0
Viewed 68

I've a function that reads N lines from a text file and save them into a first array (array_slave). After that, I'm trying to save the last for element of this array into another array (char *last_4_samples[128]). First question: is it right? Am I doing it right?

char *retrieve_last_4_samples(char slave[], char file_path[]){

        FILE *fp;
        char *line = NULL;
        size_t len = 0;
        ssize_t read;
    
        fp = fopen(file_path, "r");
    
        if (fp == NULL)
            exit(EXIT_FAILURE);
    
        char *array_slave[128];
        int i = 0;
        while ((read = getline(&line, &len, fp)) != -1)
        {
            if (strstr(line, slave) != NULL)
            {
                array_slave[i] = malloc(sizeof(*line)*len);
                strcpy(array_slave[i], line);
                i++;
            }
        }
    
        int number_of_samples = array_length(array_slave);
    
        char *last_4_samples[4];
        
        for(int i=0;i<4;i++){
            *(last_4_samples + i) = *(array_slave + i); 
        }
    
        for(int i=0;i<number_of_samples;i++)
            free(array_slave[i]);
            
    
        return last_4_samples;
    
    }

Now, in the main function, I'm trying to assign the pointer last_4_samples to another pointer, so that I can use it into the main.

char *ptr[128];
ptr = retrieve_last_4_samples(SLAVE_1,FILE_PATH);

main.c:50:9: error: assignment to expression with array type
   50 |     ptr = retrieve_last_4_samples(SLAVE_1,FILE_PATH);

What am I doing wrong?

pointers c
1 Answers

There are at least two mistakes:

  1. free(array_slave[i]); destroys memory for all elements of the array, including last 4 samples. So, whatever is in the last_4_samples is destroyed as well, because you used pointers in assignments: *(last_4_samples + i) = *(array_slave + i);

As a result the last last_4_samples becomes invalid even before return.

  1. you are returning last_4_samples which is declared as a local array on the function stack. This will become invalid in any case after return from the function.

To follow your scheme you need to

  1. duplicate your string, e.g., *(last_4_samples + i) = strdup(*(array_slave + i))
  2. avoid declaring arrays for the return: char **last_4_samples = malloc(sizeof(char*), 4);

now you can return the value of the last_4_samples pointer. Of course, you will need to free it later, but in two steps:

for(int i = 0; i < 4; i++)
   free(returned_samples[5]);
free(returned_samples);
userOAN40
Related