Thread: File information into array of structures

  1. #1
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    File information into array of structures

    Once a file is read, how would you store the information into an array of nine elements where each element is a structure named data_t.

  2. #2
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    You are going to have to provide more information.
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    I need to store information from a file that contains stats of ten different countries, and store that information in an array of ten elements, where each element is a structure named data_t.

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    What have you done so far?
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    I defined a structure and called the file for reading, but I am confused about how to store the information in the fscanf portion of the code.
    Code:
    typedef struct 
    {
    int ranking;
    char name[20];
    char state[2];
    float population, percent;
    } city_t[9];

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    Quote Originally Posted by QCMontrealQC View Post
    I defined a structure and called the file for reading, but I am confused about how to store the information in the fscanf portion of the code.
    Code:
    typedef struct 
    {
    int ranking;
    char name[20];
    char state[2];
    float population, percent;
    } city_t[9];
    The prototype for the struct should be above main(). The array of structs should be created inside main().

    Show an example of the input file, and show your attempt to read it in, please. Code must match the data to be input, if it's to work.

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    Quote Originally Posted by QCMontrealQC View Post
    I defined a structure and called the file for reading, but I am confused about how to store the information in the fscanf portion of the code.
    Code:
    typedef struct 
    {
    int ranking;
    char name[20];
    char state[2];
    float population, percent;
    } city_t[9];
    I should point out that city_t is a type that is defined as an array of 9 (not 10) struct. Are you sure that you want to define city_t as an array?

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    Code:
    for (i = 0; i < 9; i++) {
        fscanf(fp, "%d %s %s %f %f", &city_t[i].ranking, city_t[i].name, city_t[i].state, &city_t[i].population, &city_t[i].percent);
        }

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