printf format problem

i was following a tutorial at:Learn C The Hard Way A Clear & Direct Introduction To Modern C Programming

on exercise 15 i am trying the extra credit for rewriteing the program using functions.

here is my code:

Code:

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#include <stdio.h>

void indexarrayPrint();


int pointerPrint1( int *x, char **y);


int pointerPrint2( int *x, char **y);


int pointerPrint3( int *x, char **y);


int pointerPrint4( int *x, char **y);


int main(int argc, char *argv)
{
    int ages[] = {23, 43, 12, 89, 2};
    char *names[] = {
        "Alan", "Frank",
        "Mary", "John", "Lisa"
    };


    int *cur_age = ages;
    char **cur_name = names;


    indexarrayPrint(names, ages);
    pointerPrint1( cur_age, cur_name);
    pointerPrint2( cur_age, cur_name);
    pointerPrint3( cur_age, cur_name);
    pointerPrint4( cur_age, cur_name);   


    return 0;
}




void indexarrayPrint()
{
    int ages[] = {23, 43, 12, 89, 2};
    char *names[] = {
        "Alan", "Frank",
        "Mary", "John", "Lisa"
    };
   


    printf("indexarrayPrint: \n");   
    int count = sizeof(ages) / sizeof(int);
    int i = 0;
   
    for(i = 0; i < count; i++) {
        printf("%s has %d years alive.\n", names[i], ages[i]);
    }
    printf("---\n");


}


int pointerPrint1( int *x, char **y)
{
    printf("pointerPrint1: \n");
    int ages[] = {23, 43, 12, 89, 2};
   
    int count = sizeof(ages) / sizeof(int);
    int i = 0;
   
    for(i = 0; i < count; i++) {
        printf("%s is %d years old.\n", *(y+i), *(x+i));
    }
   
    printf("---\n");
}


int pointerPrint2( int *x, char **y)
{
    printf("pointerPrint2: \n");
    int ages[] = {23, 43, 12, 89, 2};
   
    int count = sizeof(ages) / sizeof(int);
    int i = 0;


    for (i = 0; i < count; i++) {
        printf("%s is %d years old again.\n", y[i], x[i]);
    }
   
    printf("---\n");
}




/* This Prints Garbadge... WIP */
int pointerPrint3( int *x, char **y)
{
    printf("pointerPrint3: \n");
    int ages[] = {23, 43, 12, 89, 2};
    char *names[] = {
        "Alan", "Frank",
        "Mary", "John", "Lisa"
    };
   


    int count = sizeof(ages) / sizeof(int);
    int i = 0;


    for (y = names, x = ages; (x - ages) < count; x++, y++) {
        printf("%s lived %d years so far.\n", y, x);
    }
   
    printf("---\n");
}


int pointerPrint4( int *x, char **y)
{
    printf("pointerPrint4: \n");
    int ages[] = {23, 43, 12, 89, 2};
   
    int count = sizeof(ages) / sizeof(int);
    int i = 0;
   
    for (i = 0; i < count; i++) {
        printf("%p lived %p years ago...\n", &y, &x);
    }
   
    printf("---\n");
}

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it works but i got 2 warnings:

Code:

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ex15_function.c: In function ‘pointerPrint3’:ex15_function.c:102:3: warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘char **’ [-Wformat]
ex15_function.c:102:3: warning: format ‘%d’ expects argument of type ‘int’, but argument 3 has type ‘int *’ [-Wformat]

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the function pointerPrint3 gives garbade output in the %s and %d areas, how do i format this to get rid of the warnings and have it print correctly?

x and y are pointers to the things you want to give printf, i.e. a pointer to an int and a pointer to a string (pointer to a char pointer). printf just wants a plain int and a plain string (char pointer). You need to dereference them to get the thing they point at. Put a * in front of each of them:

Code:

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printf("%s lived %d...", *y, *x);

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Thank You, its all good now, i will keep that in my memory for the future.

I am glad i asked, i was kinda embarrassed i was getting this kind of error.