c - for循环的退出条件是什么？

Question

我的下面的程序工作正常，但我不明白如何使用 exp[i] 作为标记位置的循环终止条件。为什么以及在什么情况下这会导致循环退出？

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

// Stack type
struct Stack
{
   int top;
   unsigned capacity;
   int* array;
};

// Stack Operations
struct Stack* createStack( unsigned capacity )
{
   struct Stack* stack = (struct Stack*) malloc(sizeof(struct Stack));

   if (!stack) 
      return NULL;

   stack->top = -1;
   stack->capacity = capacity;

   stack->array = (int*) malloc(stack->capacity * sizeof(int));

   if (!stack->array)
      return NULL;
   return stack;
}

int isEmpty(struct Stack* stack)
{
   return stack->top == -1 ;
}

char peek(struct Stack* stack)
{
   return stack->array[stack->top];
}

char pop(struct Stack* stack)
{
   if (!isEmpty(stack))
      return stack->array[stack->top--] ;
   return '$';
}

void push(struct Stack* stack, char op)
{
   stack->array[++stack->top] = op;
}


// A utility function to check if the given character is operand
int isOperand(char ch)
{
   return (ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z');
}

// A utility function to return precedence of a given operator
// Higher returned value means higher precedence
int Prec(char ch)
{
   switch (ch)
   {
      case '+':
      case '-':
         return 1;

      case '*':
      case '/':
         return 2;

      case '^':
         return 3;
   }
return -1;
}


// The main function that converts given infix expression
// to postfix expression. 
int infixToPostfix(char* exp)
{
   int i, k;

   // Create a stack of capacity equal to expression size 
   struct Stack* stack = createStack(strlen(exp));
   if(!stack) // See if stack was created successfully 
      return -1 ;

下面这个for循环的退出条件是什么？

for (i = 0, k = -1; exp[i]; ++i) 
{
    // If the scanned character is an operand, add it to output.
    if (isOperand(exp[i]))
        exp[++k] = exp[i];

    // If the scanned character is an ‘(‘, push it to the stack.
    else if (exp[i] == '(')
        push(stack, exp[i]);

    // If the scanned character is an ‘)’, pop and output from the stack 
    // until an ‘(‘ is encountered.
    else if (exp[i] == ')')
    {
        while (!isEmpty(stack) && peek(stack) != '(')
            exp[++k] = pop(stack);
        if (!isEmpty(stack) && peek(stack) != '(')
            return -1; // invalid expression             
        else
            pop(stack);
    }
    else // an operator is encountered
    {
        while (!isEmpty(stack) && Prec(exp[i]) <= Prec(peek(stack)))
            exp[++k] = pop(stack);
        push(stack, exp[i]);
    }

}

// pop all the operators from the stack
while (!isEmpty(stack))
    exp[++k] = pop(stack );

exp[++k] = '\0';
printf( "%sn", exp );
}

// Driver program to test above functions
int main()
{
   char exp[] = "a+b*(c^d-e)^(f+g*h)-i";
   infixToPostfix(exp);
   return 0;
}

Answer 1

只要 exp[i] 计算结果为 true，循环就会继续。

exp[i] 在没有任何其他上下文的情况下，计算结果为 char。可以在需要 bool 值的地方使用char。如果 char 的值为 0，则 bool 值将为 false，否则为 true。

在您的使用中，循环将继续执行 exp 的每个字符(终止空字符除外)。当遇到终止空字符时，循环将停止。

更易读的形式是使用 exp[i] != '\0'。

for (i = 0, k = -1; exp[i] != '\0'; ++i) { ... }