if语句成功后如何停止？ C程序设计

Question

I'm trying to get user input to find out deductions from salary and so they put their ID and a salary that is in range with that ID number. For example if I input ID 1 and then a salary of 500 and it should output the salary entered, salary deducted, and net salary. It does all of that, but where it checks if the salary is in range for the correct ID number it checks it with every ID number even after the ID entered has been checked.

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

int main()
{
    int ID;  // variable for ID
    float Base_Salary; // variable for base salary
    float Amount_Deducted; // variable for amount deducted from employee
    float Net_Salary; // variable for employees' net salary

    printf("%s", "Enter ID: \n"); // prompt for ID number
    scanf("%d", &ID );

    // Validate user's ID
    if (ID == 1)
        printf("You have entered the ID 1\n");
    else { if (ID == 2)
            printf("You have entered the ID 2\n");
            else { if (ID == 3)
                printf("You have entered the ID 3\n");
                    else { if (ID == 4)
                        printf("You have entered the ID 4\n");
                            else { if (ID == 5)
                                printf("You have entered the ID 5\n");
                                    else { if (ID < 1 && ID > 5)
                                        printf("You didn't enter a proper ID\n");
                                    }
                            }
                    }
            }
    }

    printf("%s", "Enter your salary in the appropriate range: \n");
    scanf("%f",&Base_Salary);

    // Validations

       if (ID == 1 && Base_Salary >= 100 && Base_Salary <= 1000) {
            printf("Base salary is in range for ID given.\n");
       }
       else {
            printf("Salary must be between 100-1000 for ID 1.\n");
       }

       if (ID == 2 && Base_Salary >= 1001 && Base_Salary <= 5000) {
            printf("Base salary is in range for ID given.\n");
       }
       else {
            printf("Salary must be between 1001-5000 for ID 2.\n");
       }

       if (ID == 3 && Base_Salary >= 5001 && Base_Salary <= 10,000) {
            printf("Base salary is in range for ID given.\n");
       }
       else {
            printf("Salary must be between 5001-10,000 for ID 3.\n");
       }

       if (ID == 4 && Base_Salary >= 10,001 && Base_Salary <= 15,000) {
            printf("Base salary is in range for ID given.\n");
       }
       else {
            printf("Salary must be between 10,001-15,000 for ID 4.\n");
       }

       if (ID == 5 && Base_Salary >= 15,001 && Base_Salary <= 20,000) {
            printf("Base salary is in range for ID given.\n");
       }
       else {
            printf("Salary must be between 15,001-20,000 for ID 5.\n");
       }

    // Calculations
    if (ID == 1) {
        Amount_Deducted = Base_Salary * 0.50;
        Net_Salary = Base_Salary - Amount_Deducted;
        printf("The Base Salary you have entered = %.2f.\n", Base_Salary);
        printf("The Amount of salary deducted is = %.2f.\n", Amount_Deducted);
        printf("The Net Salary is = %.2f.\n", Net_Salary);
    }
    if (ID == 2) {
        Amount_Deducted = Base_Salary * 1.50;
        Net_Salary = Base_Salary - Amount_Deducted;
        printf("The Base Salary you have entered = %.2f.\n", Base_Salary);
        printf("The Amount of salary deducted is = %.2f.\n", Amount_Deducted);
        printf("The Net Salary is = %.2f.\n", Net_Salary);
    }

    if (ID == 3) {
        Amount_Deducted = Base_Salary * 2.50;
        Net_Salary = Base_Salary - Amount_Deducted;
        printf("The Base Salary you have entered = %.2f.\n", Base_Salary);
        printf("The Amount of salary deducted is = %.2f.\n", Amount_Deducted);
        printf("The Net Salary is = %.2f.\n", Net_Salary);
    }

    if (ID == 4) {
        Amount_Deducted = Base_Salary * 3.50;
        Net_Salary = Base_Salary - Amount_Deducted;
        printf("The Base Salary you have entered = %.2f.\n", Base_Salary);
        printf("The Amount of salary deducted is = %.2f.\n", Amount_Deducted);
        printf("The Net Salary is = %.2f.\n", Net_Salary);
    }

    if (ID == 5) {
        Amount_Deducted = Base_Salary * 4.50;
        Net_Salary = Base_Salary - Amount_Deducted;
        printf("The Base Salary you have entered = %.2f.\n", Base_Salary);
        printf("The Amount of salary deducted is = %.2f.\n", Amount_Deducted);
        printf("The Net Salary is = %.2f.\n", Net_Salary);
    }

}

Answer 1

It really looks like switch-case is more appropriate than if-else :

// Validate user's ID
switch (ID)
{
case 1: 
    printf("You have entered the ID 1\n");
    if (Base_Salary >= 100 && Base_Salary <= 1000)
    {
        printf("Base salary is in range for ID given.\n");
    }
    Amount_Deducted = Base_Salary * 0.50;
    Net_Salary = Base_Salary - Amount_Deducted;
    break;

case 2:
    printf("You have entered the ID 2\n");
    /* Base Salary & Amount Deducted, code here. */
    break;

case 3:
    printf("You have entered the ID 3\n");
    /* Base Salary & Amount Deducted, code here. */
    break;

case 4:
    printf("You have entered the ID 4\n");
    /* Base Salary & Amount Deducted, code here. */
    break;

case 5:
    printf("You have entered the ID 5\n");
    /* Base Salary & Amount Deducted, code here. */
    break;

default:
    printf("You didn't enter a proper ID\n");
    break;
}

Answer 2

To solve your exact question, use elseif.

If an IF statement is true then all the other elseifs under it are skipped.

if (ID == 1) {
    Amount_Deducted = Base_Salary * 0.50;
    Net_Salary = Base_Salary - Amount_Deducted;
    printf("The Base Salary you have entered = %.2f.\n", Base_Salary);
    printf("The Amount of salary deducted is = %.2f.\n", Amount_Deducted);
    printf("The Net Salary is = %.2f.\n", Net_Salary);
}

else if (ID == 2) {
        Amount_Deducted = Base_Salary * 1.50;
        Net_Salary = Base_Salary - Amount_Deducted;
        printf("The Base Salary you have entered = %.2f.\n", Base_Salary);
        printf("The Amount of salary deducted is = %.2f.\n", Amount_Deducted);
        printf("The Net Salary is = %.2f.\n", Net_Salary);
    }

else if (ID == 3) {
    Amount_Deducted = Base_Salary * 2.50;
    Net_Salary = Base_Salary - Amount_Deducted;
    printf("The Base Salary you have entered = %.2f.\n", Base_Salary);
    printf("The Amount of salary deducted is = %.2f.\n", Amount_Deducted);
    printf("The Net Salary is = %.2f.\n", Net_Salary);
}

else if (ID == 4) {
    Amount_Deducted = Base_Salary * 3.50;
    Net_Salary = Base_Salary - Amount_Deducted;
    printf("The Base Salary you have entered = %.2f.\n", Base_Salary);
    printf("The Amount of salary deducted is = %.2f.\n", Amount_Deducted);
    printf("The Net Salary is = %.2f.\n", Net_Salary);
}

else if (ID == 5) {
    Amount_Deducted = Base_Salary * 4.50;
    Net_Salary = Base_Salary - Amount_Deducted;
    printf("The Base Salary you have entered = %.2f.\n", Base_Salary);
    printf("The Amount of salary deducted is = %.2f.\n", Amount_Deducted);
    printf("The Net Salary is = %.2f.\n", Net_Salary);
}

A cleaner approach would be to use arrays and use an index to loop over them, as your code is quite redundant, inside each IF block. You can easily refactor it into a method, or write it inside a FOR loop to maximize code reuse and reduce code duplication.

Answer 3

The logic says "if ID == 1 && .. check salary range ... else print error", which means for every ID that isn't 1, you will print the error.

I think you want to change this:

if (ID == 1 && Base_Salary >= 100 && Base_Salary <= 1000) {
    printf("Base salary is in range for ID given.\n");
}
else {
    printf("Salary must be between 100-1000 for ID 1.\n");
}

to this:

if (ID == 1) {
    if (Base_Salary >= 100 && Base_Salary <= 1000) {
        printf("Base salary is in range for ID given.\n");
    }
    else {
        printf("Salary must be between 100-1000 for ID 1.\n");
    }
}

and you'll have to change each of the other blocks too.

To echo some other answers, this code would be much better if you used an array to hold your salary information and a loop to iterate over the salaries, and you could get rid of all the if/else statements which are hard to read and hard to maintain.

Answer 4

Continuing from the original comment, your compiler should be telling you where your initial errors lie. If you do not warnings enabled, then get in the habit of always compiling with at least -Wall -Wextra to enable then and then read and do not accept code that compiles with any warnings. (you are unlikely to encounter any legitimate scenario where warnings cannot be eliminated) You can also add -pedantic for additional warnings if you choose.

You compiler should be telling you, eg

 salary.c:55:60: warning: left-hand operand of comma expression has no effect [-Wunused-value] if (ID == 3 && Base_Salary >= 5001 && Base_Salary <= 10, 000) { ^ 

It is even nice enough to give you a circumflex ( '^' ) pointing to the problem (eg you cannot include a comma within the number)

Secondly, as you begin programming in C, always validate ALL user input . If you fail to validate that a proper conversion has taken place in your use of scanf (by checking the return ), you can have zero confidence you are actually processing an actual value from that point forward in your code. There is nothing difficult about it, just verify that the number of expected conversions did in fact take place. You can also validate the range of the value entered at that time, eg (with a simple constant defined #define IDMAX 5 ):

printf ("Enter ID: ");  /* prompt for ID number & validate */
if (scanf ("%d", &ID) != 1 || ID < 1 || ID > IDMAX) {
    fprintf (stderr, "error: invalid ID\n");
    return 1;
}

note: it is highly unlikely that your Amount_Deducted is actually being computed by multiplying by a value greater than 1 , otherwise your Net_Salary would be negative .

Finally, while there are many ways to validate the range and deduction rate, in cases similar to this, a simple lookup table created out of arrays, provides are very efficient solution opposed to a myriad of if ... else ... or switch statements. For example:

#include <stdio.h>

#define IDMAX 5

int main (void)   /* affirmatively indicate no arguments expected */
{
    int ID = 0,   /* initialize variables */
        range [][2] = {{   100,  1000}, /* salary range lookup */
                       {  1001,  5000},
                       {  5001, 10000},
                       { 10001, 15000},
                       { 15001, 20000}};
    float Base_Salary = 0.0,
        Amount_Deducted = 0.0,
        Net_Salary = 0.0,
        rate[] = { .05, .15, .25, .35, .45 };  /* rate lookup */

    printf ("Enter ID: ");  /* prompt for ID number & validate */
    if (scanf ("%d", &ID) != 1 || ID < 1 || ID > IDMAX) {
        fprintf (stderr, "error: invalid ID\n");
        return 1;
    }

    printf ("Enter your salary in the appropriate range: ");
    if (scanf ("%f", &Base_Salary) != 1) { /* validate entry */
        fprintf (stderr, "error: invalid Base_Salary\n");
        return 1;
    }

    /* validate Base_Salary in range */
    if (Base_Salary < range[ID-1][0] || Base_Salary > range[ID-1][1]) {
        fprintf (stderr, "error: Salary must be between %d-%d for ID %d.\n",
                range[ID-1][0], range[ID-1][1], ID);
        return 1;
    }
    else
        printf ("Base salary is in range for ID given.\n");

    Amount_Deducted = Base_Salary * rate[ID-1]; /* calculations */
    Net_Salary = Base_Salary - Amount_Deducted;

    printf ("The Base Salary you have entered = %.2f.\n", Base_Salary);
    printf ("The Amount of salary deducted is = %.2f.\n",
            Amount_Deducted);
    printf ("The Net Salary is = %.2f.\n", Net_Salary);

    return 0;   /* main is type 'int' and returns a value */
}

Example Use/Output

$ ./bin/salary
Enter ID: 1
Enter your salary in the appropriate range: 99
error: Salary must be between 100-1000 for ID 1.

$ ./bin/salary
Enter ID: 1
Enter your salary in the appropriate range: 1000
Base salary is in range for ID given.
The Base Salary you have entered = 1000.00.
The Amount of salary deducted is = 50.00.
The Net Salary is = 950.00.

$ ./bin/salary
Enter ID: 2
Enter your salary in the appropriate range: 1000
error: Salary must be between 1001-5000 for ID 2.

Always compile with warnings enabled, and do not except any code until it compiles without warning, and your debug time will be greatly reduced. There are a few additional pointers included in the code comments as well. Let me know if you have any questions.

final note: While not an error, the standard style for C generally avoids Mixed_Case and caMelCase variable names in favor of all lower-case and reserving upper-case for macros and preprocessor defines. Your variable names have been left as-is to help you follow along. This is a nit, but it is worth being aware of.