有人能告诉我为什么函数没有返回值吗？ 新来的 C

Question

这段代码只是想得到收入，支出，然后在第三个function中做一些计算。我不确定问题出在哪里，当我运行它时，我一直得到值0.00。 这是我和一个朋友正在做的一个个人项目，目的是帮助获得一些 C 的经验。它可能与 switch 语句有关吗？

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

struct expense 

{ 
char name[50]; 
float amount; 
}; 

struct income
{
    char name[50];
    float amount;
};

float getExpense(void); 
float getIncome(void); 
float budgetCalc(float sum_exp, float sum_inc);

int main() 

{ 
    float sum_exp, sum_inc;
    int option; 
    printf("\t\t**************************************************************************\n");
    printf("\t\t*     Thank you for choosing Ford$ as your personal budget tracker!      *\n");
    printf("\t\t*                                                                        *\n");
    printf("\t\t*                                                                        *\n");
    printf("\t\t* The goal of this program is to allow you track your finances! You may  *\n");
    printf("\t\t* enter your specified expenses, income streams. The amount for each of  *\n");
    printf("\t\t* the expenses and income. Then, the results will be displayed. You will *\n");
    printf("\t\t* be brought back to the selection-screen after you have entered your    *\n");
    printf("\t\t* data. From the selection-screen, if you are finished, you may quit the *\n");
    printf("\t\t* program if you are finished by selecting 0. The first feature will     *\n");
    printf("\t\t* give you the option to enter your income. The second feature is the    *\n");
    printf("\t\t* expenses feature. The third will do some calculations with the given   *\n");
    printf("\t\t* details but only after you have entered your data from the first and   *\n");
    printf("\t\t* the second feature!                                                    *\n");
    printf("\t\t*                                                                        *\n");
    printf("\t\t*                   Made by ****** ******* & ***** ****                  *\n");
    printf("\t\t*                                                                        *\n");
    printf("\t\t*                                                                        *\n");
    printf("\t\t*                                                                        *\n");    
    printf("\t\t*                   Please select a feature from below:                  *\n");
    printf("\t\t*                                                                        *\n");
    START:printf("\t\t*********************************Features**********************************\n\n\t\t\t\t    \t\t");
    printf("(1) Track Income \n\t\t\t\t\t\t");
    printf("(2) Track Expenses \n\t\t\t\t\t\t");
    printf("(3) Calculator \n\t\t\t\t\t\t");
    printf("(0) Exit \n\n\t\t\t\t\t\t");
    printf("Enter: ");
    scanf("%d", &option);

    switch(option){
        case 1:
            getIncome();
            goto START;
        case 2:
            getExpense();
            goto START;
        case 3:
            budgetCalc(sum_exp, sum_inc);
            goto START;
        case 0:
            printf("\n\t\t\t\t\t\tThank you for choosing Ford$!");
            printf("\n\t\t\t\t\t\tNOW EXITING PROGRAM.....");
            exit;
    }
    
    return 0; 

} 

这为我们提供了用户输入的费用。 使用 struct expense 来存储值和数组，这和 income 都可以正常工作。

float getExpense(){ 

struct expense arr_expense[1]; 

    int i; 
    int num_exp;  
    float sum_exp=0; 

    printf("\t\t\t\t\t\tHow many expenses do you have? "); 
    scanf("%d", &num_exp); 

    for(i = 0; i < num_exp; i++ ) 
    { 
        printf("\n\t\t\t\t\t\tEnter details of Expense %d\n\n", i+1); 
        printf("\t\t\t\t\t\tEnter name of the bill: "); 
        scanf("%s", arr_expense[i].name); 
        printf("\t\t\t\t\t\tEnter amount of payment: "); 
        scanf("%f", &arr_expense[i].amount); 
    } 

    printf("\n\t\t\t\t\t\tName:\t\tAmount:\t\n"); 

    for(i = 0; i < num_exp; i++ ) 
    { 
        printf("\t\t\t\t\t\t%s\t\t%.2f\n", 
        arr_expense[i].name, arr_expense[i].amount); 
    } 

    for(i = 0; i < num_exp; i++) 
    { 
        sum_exp = sum_exp + arr_expense[i].amount; 
    } 

    printf("\n\t\t\t\t\t\tTotal expenses are: $%.2f\n\n", sum_exp); 

    printf("\t\t\t\t\t\tNOW RETURNIG TO SELECTION-SCREEN.....\n");

    return sum_exp;

} 

这给了我们用户输入的收入。 使用 struct income 将收入存储在数组中，然后在最后显示结果。

float getIncome(){

    struct income arr_income[1];
    int i; 
    int num_inc;  
    float sum_inc=0; 

    printf("\t\t\t\t\t\tHow many forms of income do you have? "); 
    scanf("%d", &num_inc); 

    for(i = 0; i < num_inc; i++ ) 
    { 
        printf("\n\t\t\t\t\t\tEnter details of Income %d\n\n", i+1); 
        printf("\t\t\t\t\t\tEnter name of the Income stream: "); 
        scanf("%s", arr_income[i].name); 
        printf("\t\t\t\t\t\tEnter the amount paid to you: "); 
        scanf("%f", &arr_income[i].amount); 
    } 

    printf("\n\t\t\t\t\t\tName:\t\tAmount:\t\n"); 

    for(i = 0; i < num_inc; i++ ) 
    { 
        printf("\t\t\t\t\t\t%s\t\t%.2f\n", 
        arr_income[i].name, arr_income[i].amount); 
    } 

    for(i = 0; i < num_inc; i++) 
    { 
        sum_inc = sum_inc + arr_income[i].amount; 
    } 

    printf("\n\t\t\t\t\t\tTotal Income is: $%.2f\n\n", sum_inc); 

    printf("\t\t\t\t\t\tNOW RETURNIG TO SELECTION-SCREEN.....\n");

    return sum_inc;
} 

这个 function 使用给定的数据运行一些计算。 目前，它只是应该显示收入的价值。 但这里唯一显示的是值 0.00 而不是实际值。 我认为它与功能和传递有关，但我看不出哪里出了问题。

float budgetCalc(float sum_inc, float sum_exp){
    printf("\t\t\t\t\t\tYou earn $%.2f! ", sum_inc);
}

Answer 1

在getIncome中，您创建一个大小为 1 的数组：

struct income arr_income[1];

然后尝试遍历num_inc数组元素。 这会写入数组末尾，调用未定义的行为。

您应该在读取num_inc的值后定义数组，然后使用该值作为数组的大小。

scanf("%d", &num_inc);
struct income arr_income[num_inc];

getExpense function 也有同样的问题。

此外，您没有使用这两个函数的返回值。 您需要将它们的返回值分配给main中的关联变量，以便它们具有有意义的值。

    case 1:
        sum_inc = getIncome();
        goto START;
    case 2:
        sum_ext = getExpense();
        goto START;

作为旁注，这不是使用goto的好地方。 请改用循环。

Answer 2

这个数组： struct income arr_income[1]; 用 1 个元素定义，稍后尝试访问超过 1 个元素，导致您的程序调用未定义的行为。

更改此：（此处和getExpense()中的类似内容）

struct income arr_income[1];
int i; 
int num_inc;  
float sum_inc=0; 

printf("\t\t\t\t\t\tHow many forms of income do you have? "); 
scanf("%d", &num_inc); 

为此：使用可变长度数组

int i; 
int num_inc;  
float sum_inc=0; 

printf("\t\t\t\t\t\tHow many forms of income do you have? "); 
scanf("%d", &num_inc); 
struct income arr_income[num_inc] ={0};//correctly sized VLA (array) of struct. 

“有人能告诉我为什么函数没有返回值吗？”
它们是，但您的代码没有读取返回值。
该段缺少左值：

switch(option){
    case 1:
        getIncome();// this function returns a value, but value is lost
        goto START;
    case 2:
        getExpense();//ditto
        goto START;

将其更改为：

switch(option){
    case 1:
        sum_inc = getIncome();//return value is captured in sum_inc
        goto START;
    case 2:
        sum_exp = getExpense();//return value is captured in sum_exp 
        goto START;

最后，关于这个部分：

    printf("\n\t\t\t\t\t\tEnter details of Expense %d\n\n", i+1); 
    printf("\t\t\t\t\t\tEnter name of the bill: "); 
    scanf("%s", arr_expense[i].name); 
    printf("\t\t\t\t\t\tEnter amount of payment: "); 
    scanf("%f", &arr_expense[i].amount); 

这里有一些使用更灵活的格式化技术替换\t字符的提示和建议。