Digit Counting in C [closed] - c

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I can't find out the problem, it doesn't give me the right answer, for example: I put 1234567890, it gives me a series of strange number, I'm new so I cant post a picture:
#include <stdio.h>
int main()
{
int n,i;
int m[10];
while((n=getchar())!='\n') {
++m[n-'0'];
}
for(i=0;i<10;++i) {
printf("%d\n",m[i]);
}
return 0;
}

Answer 1

The basic problem is that you do not initialize the array
before counting digits.
int m[10] = {0};
Also you should handle non digit values so that the program
doesn't crash e.g.
while((n=getchar())!='\n')
{
if ( isdigit(n) )
{
++m[n-'0'];
}
}
( isdigit() is available if you include ctype.h )

Answer 2

Initialize your array contents to 0 instead of whatever randomly happens to be in memory, using:
int m[10]={0};

Answer 3

A few folks have already suggested including
int m[10] = {0};
This will work. Depending upon your compiler and optimization options it may not lead to the "best" code. An alternative, that I prefer is to invoke memset() to initialize the array to 0's before using it. Here is the modified program (also note that you do not need both n and i but an optimizing compiler would take care of that too):
#include <stdio.h> /* for getchar() and printf() */
#include <string.h> /* for memset() */
#include <ctype.h> /* for isdigit() */
int main(int argc, char* argv[])
{
int n, m[11];
memset(m, 0, sizeof(m)); /* initialize counters to zero's */
while ((n=getchar()) != '\n')
{
if (isdigit(n))
{
++(m[n-'0']);
}
else
{
++(m[10]);
}
}
for(n=0;n<10;n++)
printf("%c : %d\n",'0'+n, m[n]);
printf("Filtered characters: %d\n",m[10]);
return 0;
}
A sample run looks like this:
$ ./a.out
At 12:45 p.m. I ran to the 7-11 for a 48oz big gulp
0 : 0
1 : 3
2 : 1
3 : 0
4 : 2
5 : 1
6 : 0
7 : 1
8 : 1
9 : 0
Filtered characters: 42
$
As a final note, I looked at the generated assembly instructions output from gcc for different size m[]. Gcc generates inline code (sometimes looping and sometimes unrolled depending upon optimizations) for small sized arrays (say m[10] = {0};) and generates invocations of memset() as I have shown for larger sized arrays (say m[100] = {0};). A conclusion that can be drawn from this is that leaving the code as m[10] = {0}; allows the compiler to choose the best solution based upon what it knows about the target system which is almost certainly more than what you know.

Answer 4

Well since you didnt list what your problem is, I can only guess, but without trying to compile this, the most glaring issue is the buffer overflow on the "m" array. [Something which was wrong pointed out in the comments]
Edit
Also if you enter characters other than a number or newline, you'll overwrite memory outside the "m" array too.