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Help checking a character in a structure

This is a discussion on Help checking a character in a structure within the C Programming forums, part of the General Programming Boards category; I am currently writing a function that checks whether a character string contains an "F" or "M". I have tried ...

  1. #1
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    Help checking a character in a structure

    I am currently writing a function that checks whether a character string contains an "F" or "M". I have tried strcmp() and a if statement. Here is the code I have tried, any help would be much appreciated.
    Code:
    void printFemale(Employee s[], int num) {
        int i;
        char *F;
        char female[1];
        female[0] = 'F';
        *F = female[0];
         printf( "The Female Employees Are\n");
         for( i = 0; i < num; ++i )
        {
        Trying this statement --->   if (s[i].sex[i] == 'F')
    
        Also have tried ----->    if (strcmp(s[i].sex, *F) == 0){
                printf("");
                printf("%.2f\n", s[i].pay);
            }
                        
        }
    }
    The code for the struct is as follows:
    Code:
    typedef struct Employee
    {
        char first[8];
        char initial[2];
        char last[10];
        Address ad;
        int age;
        char sex[2];
        int years;
        double pay;
    }Employee;
    I have a number of other structures that work, but when checking the character in sex[2] I just seem to be missing something.

    Thank you again.

  2. #2
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    I commented your code to show the different problems with what you tried.
    Code:
    void printFemale(Employee s[], int num) {
        int i;
        char *F;  // this is a char pointer, but it doesn't point anywhere meaningful, it points to garbage
        char female[1];  // this doesn't need to be an array, a single char would work
        female[0] = 'F';  // you set the one char in the array to 'F' -- this is okay
        *F = female[0];  // you assign the char pointed to by F (in garbage memory) to 'F', from the array -- this will likely seg fault or cause other nasty problems
         printf( "The Female Employees Are\n");
         for( i = 0; i < num; ++i )
        {
        Trying this statement --->   if (s[i].sex[i] == 'F')  // if i is not 0 or 1, you go out of bounds of the sex array.  perhaps s[i].sex[0] would work better
     
        Also have tried ----->    if (strcmp(s[i].sex, *F) == 0){  // F is a char pointer.  if F actually pointed somewhere useful, *F would be the single char it pointed to, which is not what strcmp wants, it wants a char pointer
    
                printf("");
                printf("%.2f\n", s[i].pay);
            }
                         
        }
    }
    In your case, I would do
    Code:
    if (s[i].sex[0] == 'F')
    // or
    if (strcmp(s[i].sex, "F") == 0)
    Actually, if you really only intend to store a single character for the sex field in your struct, don't use an array/string, it just confuses you. Just make it char sex;

  3. #3
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    I tried that before, the if statement which is

    Code:
    void printFemale(Employee s[], int num) {
        int i;
        
        
         printf( "The Female Employees Are\n");
         for( i = 0; i < num; ++i )
        {
            if (s[i].sex[0] == 'F'){
                printf("HELLO\n");
            }
    
        }
    }
    I think that there might be something wrong with the way the structure is set up. I have been playing around with this for hours.

  4. #4
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    Any ideas?

  5. #5
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    Quote Originally Posted by WranglerYJ View Post
    Any ideas?
    My idea is that you are not calling the printFemale function correctly.

    Tim S.

    Code:
    #include <stdio.h>
    
    typedef struct Employee
    {
        char first[8];
        char initial[2];
        char last[10];
    /*    Address ad; */
        int age;
        char sex[2];
        int years;
        double pay;
    }Employee;
    
    void printFemale(Employee s[], int num);
    
    int main()
    {
        Employee empList[1] = {{"Test", "T","Test", 21, "F", 1, 50.00}};
    
        printFemale(empList,1);
        return 0;
    }
    
    void printFemale(Employee s[], int num) {
        int i;
    
    
         printf( "The Female Employees Are\n");
         for( i = 0; i < num; ++i )
        {
            if (s[i].sex[0] == 'F'){
                printf("HELLO\n");
            }
    
        }
    }
    Last edited by stahta01; 05-11-2012 at 07:49 PM.
    "Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the universe trying to produce bigger and better idiots. So far, the Universe is winning." Rick Cook

  6. #6
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    Something wrong with the parameters of the call? I thought it was supposed to be the structure and the then the integer in the call, but when I try Employee[] it says type not allowed and I have played with other ways of using something other than employ. Am i on the right track, or is there something else to consider?

    Thank you

  7. #7
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    Read my prior post again; I updated it with good test code.

    After that, restate your question because I could not understand it.

    Edit2: "Employee" is a type not a variable name; you can NOT pass types to functions.
    Edit: You might need to review arrays http://www.cprogramming.com/tutorial/c/lesson8.html

    Tim S.
    Last edited by stahta01; 05-11-2012 at 08:03 PM.
    "Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the universe trying to produce bigger and better idiots. So far, the Universe is winning." Rick Cook

  8. #8
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    So you created an array that contains an "F", and then I can use that to check it against s[i].sex[0]? would i use strcmp. Brand new to programming so I'm sorry if I don't make myself clear. Really trying to understand this.

    Thanks again.

  9. #9
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    I created an array because your function needed one.
    The array contained elements of type Employee. The element had sex initialize to "F".
    NOTE: As suggested before (by anduril462), you should make sex a char type NOT char array.

    http://www.cprogramming.com/tutorial/c/lesson7.html

    Edit: I suggest you do all of the C tutorial to lesson 8.

    Tim S.
    Last edited by stahta01; 05-11-2012 at 08:09 PM.
    "Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the universe trying to produce bigger and better idiots. So far, the Universe is winning." Rick Cook

  10. #10
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    So inside of the function i created (well you did)
    Code:
    void printFemale(Employee s[], int num) {
        int i;
        Employee empList[1] = {{"F"}};
     
       
        
         printf( "The Female Employees Are\n");
         printf("%c", empList[0]);
         for( i = 0; i < num; ++i )
        {
            if (s[i].sex[0] == 'F'){
                printf("HELLO\n");
            }
    
        }
    }
    I read through the tutorial, and from your NOTE: in the struct declaration you two are saying to change char sex[2] to just char sex;. If I do that a lot of other things wont work. Is there a was I can :
    Code:
    if (strcmp(s[i].sex[0], emplist[0]) == 0)
         print();
    Thanks again for your help, I really appreciate it

  11. #11
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    Help Please

  12. #12
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    Don't bump your thread (read this: forum guidelines, rule #3). We'll get to it when we can -- we volunteer in our spare time after all, and we do have lives of our own .

    I gave you the strcmp solution in my original post. Also, look at your initialization of empList:
    Code:
    Employee empList[1] = {{"F"}};
    That is incorrect, and not what Tim posted.
    stahta01 likes this.

  13. #13
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    Below is code that does not initial the array; instead it assigns the value after array creation.

    Note: I am not likely to help you anymore; you are too much near the beginning of learning C; I learned C decades ago. I really do not remember enough on how to teach the first 6 weeks of learning programming. Just giving you code like I did in this post is not the best teaching method.

    Tim S.

    Code:
    int main()
    {
        Employee empList[1];
    
        strcpy(empList[0].first, "first" );
        empList[0].initial[0] = 'I';
        strcpy(empList[0].last, "last" );
        empList[0].age = 21;
        empList[0].sex[0] = 'F';
        empList[0].pay = 50.00;
    
        printFemale(empList,1);
        return 0;
    }
    YOU NEED to follow and read the information on the links posted!

    strcpy - C++ Reference
    http://www.cprogramming.com/tutorial/c/lesson9.html
    Last edited by stahta01; 05-12-2012 at 09:10 AM.
    "Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the universe trying to produce bigger and better idiots. So far, the Universe is winning." Rick Cook

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