In Go Lang 4. Program stack. For the following code, answer the following questions. Assume we are putting everything for our function calls on the stack. · Show what a stack frame/activation record for main() looks like · Show what the stack frame/activation record for the 2nd call to ctTarg look like? (Yes, this means you can skip the other stack frames) · We note that targ does not change value in any recursive call. Why doesn't the compiler just store targ once in one block of memory big enough to store a string? int ctTarg(string* list, int len, string targ) { if (len <= 0) return 0; if (*list == targ) return 1 + ctTarg(list + 1, len - 1, targ); return ctTarg(list + 1, len - 1, targ); } int main() { string pets[] = {"cat", "dog", "mouse", "cat"}; cout << ctTarg(pets, 4, "cat") << endl; }
In Go Lang
4. Program stack.
For the following code, answer the following questions. Assume we are putting everything for our function calls on the stack.
· Show what a stack frame/activation record for main() looks like
· Show what the stack frame/activation record for the 2nd call to ctTarg look like? (Yes, this means you can skip the other stack frames)
· We note that targ does not change value in any recursive call. Why doesn't the compiler just store targ once in one block of memory big enough to store a string?
int ctTarg(string* list, int len, string targ) {
if (len <= 0) return 0;
if (*list == targ)
return 1 + ctTarg(list + 1, len - 1, targ);
return ctTarg(list + 1, len - 1, targ);
}
int main()
{
string pets[] = {"cat", "dog", "mouse", "cat"};
cout << ctTarg(pets, 4, "cat") << endl;
}
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