Printf不显示并返回值3221225477

Question

New to C and programming in general, so I am having kind of a tough time with structs when combined with arrays and pointers. I'm trying to create a struct with attempts, then create an array pointer (towards the struct) repeating it 10 times. Then find the average for every struct and print it.

Everything seems to work normally providing a return value of 0 until the loop.

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

typedef struct Tries {
float attempts1;
float attempts2;
float attempts3;
float aver;
}Try;

int main(int argc, char *argv[]) {
int i,size=10,at1,at2,at3;
Try** arrayofTries= malloc (sizeof(Try)*size);
for (i=0;i<size;i++){

        arrayofTries[i]->attempts1= rand () %(900 - 700)+700;
        arrayofTries[i]->attempts2= rand () %(900 - 700)+700;
        arrayofTries[i]->attempts3= rand () %(900 - 700)+700;
        at1= arrayofTries[i]->attempts1;
        at2= arrayofTries[i]->attempts2;
        at3=arrayofTries[i]->attempts3;
        arrayofTries[i]->aver = (at1+at2+at3)/3;
     printf ("The average of %d person is%f",i,arrayofTries[i]->aver);
}

return 0;
}

Answer 1

arrayofTries should be of type Try * not Try **

Try *arrayofTries = malloc (sizeof(Try) * size);

So all your -> should be simple dots . .

Answer 2

You are allocating a pointer to pointer to Try but not allocating the direct pointer to the elements themselves. It looks like your intent is to declare a 1D array, not a 2D array, so what you should do is change the type of arrayofTries from Try ** to Try * .

Also, when you're done using memory allocated with malloc , free it:

free(arrayofTries);
arrayofTries = NULL;