【发布时间】:2017-09-06 09:41:05
【问题描述】:
我需要你的帮助。我想将我的代码输出导出到文本文件中。我真的不知道如何处理它。有人可以帮我解决这个问题。非常感谢
这是我的代码:
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
struct node {
int data;
struct node *next;
};
struct node *top = NULL;
struct node* createNode(int data){
struct node *p = (struct node *) malloc(sizeof (struct node));
p->data = data;
p->next = NULL;
}
void push (int data){
struct node *ptr = createNode(data);
if (top == NULL){
top = ptr;
return;
}
ptr->next = top;
top = ptr;
}
int pop(){
int data;
struct node *temp;
if (top == NULL)
return -1;
data = top->data;
temp = top;
top =top->next;
free(temp);
return (data);
}
int main(){
char str[100];
int i = 0, data = -1, operand1, operand2, result;
printf("Expression in postfix format: ");
fgets(str, 99, stdin);
for (; i < strlen(str); i++){
if (isdigit(str[i])){
data = (data == -1) ? 0 : data;
data = (data * 10) + (str[i] - 48);
continue;
}
if (data != -1){
push(data);
}
if (str[i] == '+' || str[i] == '-' || str[i] == '*' || str[i] == '/'){
operand2 = pop();
operand1 = pop();
if (operand1 == -1 || operand2 == -1)
break;
switch (str[i]){
case '+':
result = operand1 + operand2;
push(result);
break;
case '-':
result = operand1 - operand2;
push(result);
break;
case '*':
result = operand1 * operand2;
push(result);
break;
case '/':
result = operand1 / operand2;
push(result);
break;
}
}
data = -1;
}
if (top != NULL && top->next == NULL)
printf("Postfix Evaluation: %d\n", top->data);
else
printf("Invalid Expression!\n");
return 0;
}
【问题讨论】:
-
您需要将 printfs 替换为 fprintfs
-
prog > output.txt -
或使用
freopen -
贴出的代码编译不干净!。除其他外,它缺少以下声明:
#include <ctype.h>for theisdigit() -
函数:
createNode()签名表示它返回指向struct node的指针。但是,该函数中不存在这样的 return 语句。