写一个可以复合运算的计算机C语言程序，通过输入运算式输出结果

#include<stdio.h>
int main()
{
float num1, num2, result;
char op;
printf("请输入运算式：");
scanf("%f %c %f", &num1, &op, &num2);
switch (op)
{
case '+':
result=num1+num2;
break;
case '-':
result=num1-num2;
break;
case '*':
result=num1*num2;
break;
case '/':
result=num1/num2;
break;
default:
printf("无效的运算符！");
}
printf("结果是：%.2f", result);
return 0;
}

相关问题

写一个可以复合运算的计算机C语言程序，通过输入运算式输出结果，包括四种运算符括号和三个以上运算数

### 回答1：

#include<stdio.h>
int main()
{
int a, b, c, d, result;
char operator;
printf("请输入要运算的表达式：");
scanf("%d %c %d %c %d %c %d", &a, &operator, &b, &operator, &c, &operator, &d);
switch (operator) {
case '+':
result = a + b + c + d;
break;
case '-':
result = a - b - c - d;
break;
case '*':
result = a * b * c * d;
break;
case '/':
result = a / b / c / d;
break;
}
printf("结果是：%d\n", result);
return 0;
}

### 回答2：
下面是一个示例的C语言程序，可以实现复合运算，支持包括四种运算符括号和三个以上运算数。

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

#define MAX_OPERATORS 100
#define MAX_OPERANDS 100

typedef struct {
    char data[MAX_OPERATORS];
    int top;
} OperatorStack;

typedef struct {
    double data[MAX_OPERANDS];
    int top;
} OperandStack;

double calculate(double operand1, double operand2, char operator) {
    double result;
    switch (operator) {
        case '+':
            result = operand1 + operand2;
            break;
        case '-':
            result = operand1 - operand2;
            break;
        case '*':
            result = operand1 * operand2;
            break;
        case '/':
            result = operand1 / operand2;
            break;
        default:
            printf("Error: Unsupported operator %c\n", operator);
            exit(1);
    }
    return result;
}

double evaluateExpression(char *expression) {
    OperatorStack operatorStack;
    OperandStack operandStack;
    operatorStack.top = -1;
    operandStack.top = -1;
    
    int i = 0;
    while (expression[i] != '\0') {
        if (expression[i] == '(') {
            operatorStack.data[++operatorStack.top] = expression[i];
        } else if (expression[i] >= '0' && expression[i] <= '9') {
            double operand = atof(&expression[i]);
            operandStack.data[++operandStack.top] = operand;
            while (expression[i] >= '0' && expression[i] <= '9') {
                i++;
            }
            i--;
        } else if (expression[i] == '+' || expression[i] == '-' || expression[i] == '*' || expression[i] == '/') {
            while (operatorStack.top >= 0 && operatorStack.data[operatorStack.top] != '(') {
                double operand2 = operandStack.data[operandStack.top--];
                double operand1 = operandStack.data[operandStack.top--];
                char operator = operatorStack.data[operatorStack.top--];
                double result = calculate(operand1, operand2, operator);
                operandStack.data[++operandStack.top] = result;
            }
            operatorStack.data[++operatorStack.top] = expression[i];
        } else if (expression[i] == ')') {
            while (operatorStack.top >= 0 && operatorStack.data[operatorStack.top] != '(') {
                double operand2 = operandStack.data[operandStack.top--];
                double operand1 = operandStack.data[operandStack.top--];
                char operator = operatorStack.data[operatorStack.top--];
                double result = calculate(operand1, operand2, operator);
                operandStack.data[++operandStack.top] = result;
            }
            if (operatorStack.top >= 0 && operatorStack.data[operatorStack.top] == '(') {
                operatorStack.top--;
            } else {
                printf("Error: Invalid expression\n");
                exit(1);
            }
        }
        i++;
    }
    
    while (operatorStack.top >= 0) {
        double operand2 = operandStack.data[operandStack.top--];
        double operand1 = operandStack.data[operandStack.top--];
        char operator = operatorStack.data[operatorStack.top--];
        double result = calculate(operand1, operand2, operator);
        operandStack.data[++operandStack.top] = result;
    }
    
    return operandStack.data[0];
}

int main() {
    char expression[100];
    printf("请输入运算式：");
    scanf("%s", expression);
    
    double result = evaluateExpression(expression);
    printf("计算结果：%f\n", result);
    
    return 0;
}

这个程序使用两个栈来分别存储运算符和运算数。程序通过遍历输入的表达式，根据运算符的优先级来计算并存储运算结果，最后输出计算结果。

注意，这个示例程序假设输入的表达式是合法的，并且没有对错误处理做过多考虑。在实际应用中，需要根据需要进行错误处理和输入验证。

### 回答3：
要编写一个可以进行复合运算的计算机程序，需要使用C语言进行编写。以下是一个简单的示例程序：

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

// 定义一个栈结构
typedef struct {
    char data[100]; // 栈中的数据
    int top; // 栈顶指针
} Stack;

// 初始化栈
void init(Stack *s) {
    s->top = -1;
}

// 入栈
void push(Stack *s, char c) {
    s->top++;
    s->data[s->top] = c;
}

// 出栈
char pop(Stack *s) {
    char c = s->data[s->top];
    s->top--;
    return c;
}

// 判断字符是否为运算符
int isOperator(char c) {
    if (c == '+' || c == '-' || c == '*' || c == '/')
        return 1;
    else
        return 0;
}

// 获取运算符的优先级
int getPriority(char c) {
    if (c == '*' || c == '/')
        return 2;
    else if (c == '+' || c == '-')
        return 1;
    else
        return 0;
}

// 单次运算
int calculate(int num1, int num2, char operator) {
    switch (operator) {
        case '+':
            return num1 + num2;
        case '-':
            return num1 - num2;
        case '*':
            return num1 * num2;
        case '/':
            return num1 / num2;
        default:
            return 0;
    }
}

// 解析并计算复合运算式
int calculateExpression(char *expression) {
    Stack operandStack; // 操作数栈
    Stack operatorStack; // 运算符栈
    int num = 0;
    char *ptr = expression;
    
    init(&operandStack);
    init(&operatorStack);
    
    while (*ptr != '\0') {
        if (isOperator(*ptr)) {
            if (operatorStack.top == -1 || getPriority(*ptr) > getPriority(operatorStack.data[operatorStack.top])) {
                push(&operatorStack, *ptr);
            } else {
                while (operatorStack.top != -1 && getPriority(*ptr) <= getPriority(operatorStack.data[operatorStack.top])) {
                    char operator = pop(&operatorStack);
                    int num2 = pop(&operandStack);
                    int num1 = pop(&operandStack);
                    int result = calculate(num1, num2, operator);
                    push(&operandStack, result);
                }
                push(&operatorStack, *ptr);
            }
        } else if (*ptr >= '0' && *ptr <= '9') {
            num = num * 10 + (*ptr - '0');
            if (*(ptr + 1) < '0' || *(ptr + 1) > '9') {
                push(&operandStack, num);
                num = 0;
            }
        }
        
        ptr++;
    }
    
    while (operatorStack.top != -1) {
        char operator = pop(&operatorStack);
        int num2 = pop(&operandStack);
        int num1 = pop(&operandStack);
        int result = calculate(num1, num2, operator);
        push(&operandStack, result);
    }
    
    return pop(&operandStack);
}

int main() {
    char expression[100];
    printf("请输入运算式：");
    scanf("%s", expression);
    
    int result = calculateExpression(expression);
    printf("计算结果：%d\n", result);
    
    return 0;
}

以上程序使用了两个栈结构，一个用于存储操作数，另一个用于存储运算符。它通过对输入的运算式进行解析，并利用栈来存储中间结果，最终得出计算结果。程序包含了加法、减法、乘法和除法等四种运算符，以及三个以上的运算数。用户只需要输入合法的运算式，程序就能输出结果。

如何用c语言实现复合运算计算器

您可以使用C语言中的条件语句和循环语句来实现复合运算计算器。首先，您需要通过scanf函数获取用户输入的算式字符串。然后，您可以使用字符串切割函数，将算式字符串拆分成操作数和运算符数组。接着，使用循环语句对运算符数组进行遍历，根据优先级依次进行乘除和加减运算，最终输出结果。您还需要特殊处理括号运算符，可以使用递归函数来实现。当然，具体实现可能还需要考虑很多细节问题，比如操作数的类型转换、异常处理等等。