Question

Consider the following snippet of code. (Assume that input strings are at most 255 characters long.)

char* to_upper_case(char* original) {
  char capstr[255];
  unsigned int i;
  for (i = 0; original[i] != '\0'; ++i) {
    if (original[i] >= 'a' && original[i] <= 'z') {
      capstr[i] = original[i] - (char)'a' + (char)'A';
    } else {
      capstr[i] = original[i];
    }
  }
  capstr[i] = '\0';
  return capstr;
}

void main() {
  printf("%s", to_upper_case("the c programming language"));
}

a) What will be the output? Trace this code manually, do not run it. (Hint: be careful!)

b) Why does the program give that output? Explain in terms of memory and pointer usage.

Answer 1

a) There will be segmentation fault when the program runs.

b)The program would generate a warning when compiled because we are trying to return a char pointer which is local to the function. When the function returns, this address is junk and will lead to segmentation fault when trying to print in main.