A program that displays the characters which appeared least/most often - c

Question

I've just written the code which is supposed to display the most/least (depending on a switch choice) appearing character. It has to work with standard input, that is from keyboard as well as with a text file. It's crucial for this program to show all char that appeared most often. For example in the input "abbcc" it should display b and c.
It runs, but it doesn't show characters. The question is: why?
Here's the code:
a) main
int main (int argc, char *argv[]) {
int string;
int allChars[256] = {0};
while ( (string=getchar())!=EOF )
allChars[string]++;
if (argc < 2)
mostOften(allChars);
else {
switch (argv[1][1]) {
case 'm': case 'M':
mostOften(allChars);
break;
case 'l': case 'L':
leastOften(allChars);
break;
default:
mostOften(allChars);
break;
} //switch
} // else
} // main
b) sf.c
void mostOften(int *s) {
int i, j;
int max[256] = {0} ;
int max_count = 1;
for (i=0; i<256; i++) {
if (s[i]) {
if (s[i] > max_count) {
for (j=0; j<256; j++)
max[j]=0;
max[i]=1;
max_count = s[i];
} // 2nd_if
else if (s[i] == max_count)
max[i]=1;
else
continue;
} // 1st_if
} //1st_for
printf("The most appearing characters are: ");
for (i=0; i<256; i++) {
if (max[i])
putchar(s[i]);
} //2nd_for
}
void leastOften(int *s) {
int i, j;
int min[256] = {0} ;
int min_count = 1000;
for (i=0; i<256; i++) {
if (s[i]) {
if (s[i] < min_count) {
for (j=0; j<256; j++)
min[j]=0;
min[i]=1;
min_count = s[i];
} // 2nd_if
else if (s[i] == min_count)
min[i]=1;
} // 1st_if
} //1st_for
printf("The least appearing characters are: ");
for (i=0; i<256; i++) {
if (min[i])
putchar(s[i]);
} //2nd_for
}
Thank you.
EDIT1: deleted the mistake in the description
EDIT2: changed from putchar(min[i]), putchar(max[i]) to putchar(s[i])

Answer 1

In your code for mostOften, i is the index to s, an that is the character itelf. So when you print out the list of characters that occur most often, you should print the character i instead of max[i]:
for (i = 0; i < 256; i++) {
if (max[i]) putchar(i);
}
Another approach that doesn't require zeroing out the whole array is to make a contiguous list of characters. The length of the list is variable and kept in a separate variable, nmax. When you find a new maximum, just set the length of the array to zero. Don't worry about the array contents; max is only guaranteed to have sensible data up to nmax:
void mostOften(int *s)
{
int max[256]; // list of characters
int nmax = 0; // actual length of max array
int max_count = 0;
int i;
for (i = 0; i < 256; i++) {
if (s[i] && s[i] >= max_count) {
if (s[i] > max_count) nmax = 0;
max[nmax++] = i;
max_count = s[i];
}
}
for (i = 0; i < nmax; i++) printf("'%c'\n", max[i]);
}

Answer 2

change
printf("The most appearing characters are: ");
for (i=0; i<256; i++) {
if (max[i])
putchar(s[i]);
} //2nd_for
to
printf("The most appearing characters are: ");
for (i=0; i<256; i++) {
if (max[i])
putchar(i);
} //2nd_for
since you generated s[i] as s[<char-value>] = <num-occurances-of-value>.