language : C

In this part you will write a calculator for arithmetic expressions in prefix notation.

------------1. Write a function that gets an array of tokens representing an arithmetic expression in the prefix notation, and computes the evaluation of the expression. Use the stack evaluation discussed in class. int prefix_eval_tokens(char** prefix_expr, int n_tokens) Examples: On input ["+", "5", "-", "7", "2"] the function will return 10. This is because the expression is : ( 5 + ( 7 - 2 ) ) = 10. * You may assume that all intermediate values in the computation are between 0 and 99 ** You may assume that the results of all division operations are integers. *** You may use the stack implementation given in the assignment. The data in stack is of type void*. Use casting if necessary. -----------------2. Use the function prefix_eval to write a function that gets a string representing an arithmetic expression in prefix notation and evaluates the corresponding expression. int prefix_eval(char* expression) * You may assume that all intermediate values in the computation are between 0 and 99 ** You may assume that the results of all division operations are integers. *** You may assume that all expression are formatted correctly. Examples: "/ 25 5" "+ 5 - 7 2" "- / + 15 13 - 5 1 + 2 4"

the basic stack functions could be use in the question are :

stack_t* stack_create() { stack_t* s = (stack_t*) malloc(sizeof(stack_t)); s->top = NULL; return s; }

// pushes a given item to the stack // Pre condition: stack is not empty void stack_push(stack_t* s, void* item) { node_t* newTop = (node_t*) malloc(sizeof(node_t)); newTop->data = item; newTop->next = s->top; s->top = newTop; }

// pops the top element from the stack // Pre condition: stack is not empty void* stack_pop(stack_t* s) { void* ret = s->top->data; node_t* prevTop = s->top; s->top = prevTop->next; free(prevTop); return ret; }

// returns 1 if the stack is empty, and returns 0 otherwise int stack_is_empty(stack_t* s) { return (s->top == NULL); }

void stack_free(stack_t* s) { while (s->top) { node_t* prevTop = s->top; s->top = prevTop->next; free(prevTop); } free(s); }