C - Declaring an array with an undefined value

Question

I am trying to determine the size of an array by having the user input the size of the array after being prompted (because I don't know how many grades I'll have per test)...

It doesnt seem to like the line where scanf ("%c",&grades[i]);

Here is the whole function:

#include <stdio.h>

 int main (void)
{
int numGrades;
char grades;
char i;
int x;


printf ("Enter The Amount Of Numbers In Your Array: ");
scanf("%i", &numGrades);/*Stores Amount Of Grades In The Array*/

for (x = 0; x < numGrades; ++x)
  {
  printf ("\nEnter the grade: ");
  scanf ("%c",&grades[i]);
  }

return 0; 
}

How can I pass the array size as a parameter so that I can accept an array of any size? (I will be adding a function that will take all the grades and combine them together by letter)

Change char grades to char* grades. And after scanf, do this. grades = malloc(numGrades); — sgarizvi
– sgarizvi, Commented Mar 7, 2013 at 6:33
grades is supposed to be an array (or a pointer to char), not a single char — uba
– uba, Commented Mar 7, 2013 at 6:33

Some programmer dude · Accepted Answer · 2013-03-07 06:52:51Z

3

You have two choices:

If your compiler support C99 variable length arrays you could declare the array after you get the size:
```
scanf("%i", &numGrades);
char grades[numGrades];
```

If not then you have to dynamically allocate the array on the heap:

scanf("%i", &numGrades);
char *grades = malloc(numGrades * (sizeof *grades));

edited Mar 7, 2013 at 6:52

answered Mar 7, 2013 at 6:34

Some programmer dude

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4 Comments

Ed Swangren Over a year ago

I would just note that VLA's created from user input are a horrible idea (if not just a horrible idea in general...). Use malloc. Also, sizeof(char) is defined to always be 1.

Jonathan Leffler Over a year ago

VLAs created on unchecked user input sizes are a bad idea. If you've checked the size for sanity (strictly positive — not zero, not negative — and not too large for some nebulous definition of 'too large', perhaps 4 KiB or 64 KiB), then there's not likely to be any harm done.

Some programmer dude Over a year ago

About the sizeof(char) thing, just future-proofing in case the OP wants to use some other type. :)

Carl Norum Over a year ago

@JoachimPileborg, then use sizeof *grades.

Carl Norum · Accepted Answer · 2013-03-07 06:33:05Z

2

You declared grades as a single char, not an array. Either move the declaration to after you read in numGrades and make it a VLA, like this:

char grades[numGrades];

Or use dynamic allocation:

char *grades = malloc(numGrades);

If you choose the latter, don't forget to free() it.

answered Mar 7, 2013 at 6:33

Carl Norum

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