我不认为我正确使用指针。 C

Question

So I've been trying to do this assignment for a few hours and I can't figure it out, I'm 99% sure it has to do with how I used the pointers. And the error I get is a break 0xC0000005: Access violation writing location 0xCCCCCCCC.which I think means some of my pointers are null. Sorry for the amateur code in advance..

#include<stdio.h>

void load(int *salary)
{
    printf("Please enter your salary: ");
    scanf("%d", *salary);
}

void calc(int *salary, float *rate, int *raise, int *newsalary)
{
    rateofsalary(&salary);
    *raise = *salary/(*rate);
    *newsalary = *raise+*salary;
}
float rateofsalary(int *salary)
{
    float rate;
    if(*salary<0 && *salary>=30000)
        rate = 7.0;
    else 
        if(*salary<30000 && *salary>=40000)
            rate = 5.5;
        else 
            if(*salary<40000)
                rate = 4.0;
    return rate;
}

void print(int *salary, float *rate, int *raise, int *newsalary)
{
    printf("|     | Salary | Rate % | Raise | New Salary |\n");
    printf("|     | %d     | %0.2f  | %d    | %d         |\n", salary, rate, raise, newsalary);
}

void main()
{
    int salary, raise, newsalary;
    float rate;
    load(&salary);
    rateofsalary(&salary);
    calc(&salary, &rate, &raise, &newsalary);
    print(&salary, &rate, &raise, &newsalary);
}

Answer 1

for the load function it should be

void load(int *salary)
{
    printf("Please enter your salary: ");
    scanf("%d", salary);
}

Answer 2

scanf("%d", *salary);

change it to

scanf("%d", salary);

and the printf function should be

printf("|     | %d     | %0.2f  | %d    | %d         |\n", *salary, *rate, *raise, *newsalary);

Answer 3

well in this line scanf("%d", *salary); instead of *salary just put salary so replace it with this scanf("%d", salary);

Answer 4

In general, your problem is that you don't know when to quit :)

If something is of type int * , that means it's a pointer to an int , and it's not necessary nor correct to turn around and use & on it, which turns it into a pointer to pointer to int. You make this mistake multiple times, not only in the first short function as other answers have pointed out, but in multiple places. You manage to turn the variable salary from main into a pointer to pointer to pointer to int by the time you're through, adding a level of indirection each time you call another function.

Just keep track of what each function wants. If a function accepts an int* as an argument, and calls another function that wants an int* , then it can just pass the variable along as-is: you don't need to add a & to it.

Answer 5

There are few places you use pointers wrong. What others didn't mention yet is: in function:

void calc(int *salary, float *rate, int *raise, int *newsalary)
{
    rateofsalary(&salary);
    *raise = *salary/(*rate);
    *newsalary = *raise+*salary;
}

you call rateofsalary with &salary , which is pointer to pointer to int (int **), but formal argument of rateofsalary is declared as int * which is pointer to int .

It's not the best idea to variable by reference to function, that shouldn't be allowed to change that variable. Pass it by value or declare function parameter as const .
Whole idea of passing by reference is to be able to modify variable passed to function inside function body and you seem to be missing what pointers actually are or practical use of them.