What is the problem with pointers in c?

Strange problem regarding C and pointers?

• I'm trying to write an encryption program that generates a random substitution map for the letters of the alphabet.The function that generates this substitution map returns a pointer and even though i'm sure the substitution map is created succesfully(by printing it inside the function) i can't use it because there's a runtime error when it's trying to loop through this new random array.But i can print the single array elements without a problem. But it doesn't work if i use a for or while loop. Here's relevant part of the program. First the function that generates the substitution map : char* generateRandomSubMap1(char *ptr) { char *ptrSub; char a[26]; int i,j,k,flag; srand(time(NULL)); for(j=0;j<=25;j++) { do { flag = 0; i = rand() % 26; for(k=0;k<j;k++) if(a[k] == *(ptr + i)) flag = 1; }while(flag); a[j] = *(ptr+i); printf(" %c",a[j]); } printf("\n"); ptrSub = a; return ptrSub; } Now in main i write ptrSubA = generateRandomSubMap1(ptrA); ( where ptrA points to the array holding the letters of the alphabet in order) while(*ptrSubA != '\0'){ printf(" %c",*ptrSubA); ptrSubA++;} or for(j=0;j<=25;j++) printf(" %c", *(ptrSubA+j)); (j is an integer) and it doesn't work. This works though : printf(" %c ",*(ptrSubA +1)); But it's of no use without a loop obviously. Any help would be really appreciated and sorry for the ugly wall of text :)

• Answer:

  Although you return a pointer to the your map, your map is allocated on the stack. The memory is only allocated while generateRandomSubMap1 is being called, when it returns the next function that gets call will reuse the memory for its own purposes, overwritting it. You need to do one of two things. If you only need one of these maps, make the "a" array static, either by simply inserting the word "static" before its definition, or by moving the whole definition outside of generateRandomSubMap1. The alternative is to allocate memory from the heap. That would change the line: char a[26]; to char* a = (char*)malloc(26); Nothing else needs to change, although ideally callers to your function would eventually call free(ptr); on the object they were returned in order to give the memory back.