Is there a way to store the value of a pointer but do not access the address?-CodePudding

Now I create a double pointer to store some words get from a .txt document (Since there is no string in C language.). I tried to use fgets and fscanf. When I run fgetsand fscanffor the first time, I can store pointers in double pointers. But when I run it for the second time, I find that the value of the first time is completely overwritten. This is a part of code.

char **out = (char **)malloc(3 * sizeof(char *));
FILE *dictionary = NULL;

dictionary = fopen(filename, "r");

char store[6];

fscanf(dictionary, "%s", store);
printf("%s\n", store);
store[5] = '\n';
out[0] = store;
strcpy(store, "");
fscanf(dictionary, "%s", store);
out[1] = store;
printf("%s\n", out[0]);
printf("%s\n", out[1]);

fclose(dictionary);
return out;

The content of txt file is

cigar
rebut
sissy

My expected result from the code above is

cigar
cigar
rebut

but the exact output from the code above is

cigar
rebut
rebut

I just tried to store their values one by one, but it seems to tamper with my previous values as well. I've replaced pointers with arrays, but this keeps showing up. What do I need to pay attention to? Or is there a better alternative?

CodePudding user response：

Here, you need to allocate memory again for every 1D character array out[0] = (char*)malloc(5 * sizeof(char)); and use strcpy, so the below is the corrected code.

char **out = (char **)malloc(3 * sizeof(char *));
FILE *dictionary = NULL;

dictionary = fopen(filename, "r");

char store[6];

out[0] = (char*)malloc(5 * sizeof(char));
fscanf(dictionary, "%s", store);
printf("%s\n", store);
strcpy(out[0], store);

out[1] = (char*)malloc(5 * sizeof(char));
fscanf(dictionary, "%s", store);
strcpy(out[1], store);
printf("%s\n", out[0]);
printf("%s\n", out[1]);

fclose(dictionary);

return 0;