User defined function - returning a pointer / string

Question

Can anyone point out the obvious? I can't seem to find what I've done wrong.

This code produces quite a bit of a pause followed by no visible output. Debug crashes so it's a real good attempt.

Edit; The intents just to understand how to output a char via a user defined function when a int is passed to it. eg you enter a test score of 85 printf should print HD. I had no idea getchar() was ASCII value, this might explain the output I was getting earlier. Thanks Bodo.

If I change the printf - %d I do get numbers so thought I was getting ASCII but they have not added up to something I've been able to decrypt. Numerical entry of 85 = 100 & 65 = 104

If it's super wrong let me know and I'll go back and make this code work without a user defined function first.

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

char determine_grade(int test_score);

int main()
{
    int test_score = 0;
    char *test_result;

    printf("enter test score:");
    test_score =getchar();

    test_result = determine_grade(test_score);

    printf("grade is: %s", test_result);

    return 0;
}

char determine_grade(int test_score)
{
    char *test_result;
        if (test_score >= 85)
        test_result = "HD";
    else if (test_score >= 75)
        test_result = "D";
    else if (test_score >= 65)
        test_result = "C";
    else if (test_score >= 55)
        test_result = "P1";
    else if (test_score >= 50)
        test_result = "P2";
    else if (test_score >= 40)
        test_result = "F1";
    else
        test_result = "F2";
    
    return test_result;
}

Your function returns a char, not a char* - also turn on compiler warnings, your compiler will clearly tell you where the problems are — UnholySheep
– UnholySheep, Commented May 10, 2021 at 9:58
Thanks, working in VS and I closed it out and re added it. Compiler then had warnings where as before it was compiling. Strange. — desmo
– desmo, Commented May 10, 2021 at 10:14
@desmo This statement test_score =getchar(); does not make a sense. Instead use scanf( "%d", &test_score ); — Vlad from Moscow
– Vlad from Moscow, Commented May 10, 2021 at 10:27
What do you expect the user to enter after enter test score:? Function getchar() will return the code (ASCII value or other encoding) of a single character or the first byte of a multi-byte character. In determine_grade you compare the value with numbers 85...40 which would correspond to ASCII characters 85 = 'U', 75 = 'K' etc. So any input that starts with, for example'U', 'V', '[', 'a' will match the first condition >= 85, input with any of 'K'...'T' will match the second condition >=75 etc. Is this what you want? Please edit your question to answer. — Bodo
– Bodo, Commented May 10, 2021 at 10:30

Vlad from Moscow · Accepted Answer · 2021-05-10 10:43:08Z

2

For starters neither declaration from these headers

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

are used in your program. So you may remove them.

The function determine_grade returns a pointer to a string literal insetead of an object of the type char, Also the parameter test_score should have an unsigned integer type. So the function should be declared like

const char * determine_grade( unsigned int test_score );

and in its definition you have also to write

const char * determine_grade( unsigned int test_score )
{
    const char *test_result;
    //...
}

Correspondingly in main you should write

unsigned int test_score = 0;
const char *test_result;

This call of getchar does not make a sense because the function returns just one internal code of a character. Instead write

scanf( "%u", &test_score );

answered May 10, 2021 at 10:43

Vlad from Moscow

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1 Comment

desmo Over a year ago

A massive thank-you Vlad and the rest of you. I'm off to learn what an unsigned int is and pull this solution apart to understand it better.

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