Iterate through a string that's on an array in C - c

Question

I have a code like this:
char* s[4][10] = { "abc\0", "aab\0", "cbb\0" };
a want to type a character, like.. 'a', and iterate through the array of strings, check on each string if there's 'a' on it, and then count how many there's on it..
In this case, what I wanted to get is "abc" : 1, "aab" : 2, "cbb" : 0 ...
how can I do the loops to get this result?

Answer 1

char input = '\0';
scanf("%c", &input);
int i, j;
for(i=0; i<4; i++)
{
int count = 0;
for(j=0;j<10;j++)
{
if(s[i][j] == input) count++;
}
printf("%s : %d", s[i], count);
}
although please note, that simply copying and pasting will not teach you anything.

Answer 2

Here is a demonstrative program that shows how you can deal with different declarations of arrays
#include <stdio.h>
int main(void)
{
char* s[4][10] = { { "abc", "aab", "cbb" } };
for ( int i = 0; i < 4; i++ )
{
for ( int j = 0; j < 10; j++ )
{
if ( s[i][j] )
{
size_t count = 0;
for ( char *p = s[i][j]; *p != '\0'; ++p )
{
if ( *p == 'a' ) ++count;
}
printf( "\"%s\": %zu\n", s[i][j], count );
}
}
}
printf( "\n" );
char* t[4] = { "abc", "aab", "cbb" };
for ( char **p = t; *p != NULL; ++p )
{
size_t count = 0;
for ( char *q = *p; *q != '\0'; ++q )
{
if ( *q == 'a' ) ++count;
}
printf( "\"%s\": %zu\n", *p, count );
}
printf( "\n" );
char u[4][10] = { "abc", "aab", "cbb" };
for ( int i = 0; i < 4; i++ )
{
if ( u[i][0] != '\0' )
{
size_t count = 0;
for ( char *p = u[i]; *p != '\0'; ++p )
{
if ( *p == 'a' ) ++count;
}
printf( "\"%s\": %zu\n", u[i], count );
}
}
return 0;
}
The program output is
"abc": 1
"aab": 2
"cbb": 0
"abc": 1
"aab": 2
"cbb": 0
"abc": 1
"aab": 2
"cbb": 0
It seems that you mean either the second variant of the array declaration or the third variant of the array declaration. The first variant looks strange.:)
Take into account that string literals already contain terminating zero. So there is no sense to write, for example
"abc\0"
It is enough to write simply
"abc"
The sizeof( "abc" ) is equal to 4 due to the presence of the terminating zero.

Answer 3

#include <stddef.h>
#include <stdio.h>
int main(int argc, char *argv[]) {
char s[10][4] = { "abc", "aab", "cbb" };
size_t i;
for (i = 0; i != sizeof(s) / sizeof(s[0]); i++) {
fprintf(stdout, "%zu: %s\n", i, s[i]);
}
return 0;
}
or
#include <stddef.h>
#include <stdio.h>
int main(int argc, char *argv[]) {
char *s[4] = { "abc", "aab", "cbb" };
size_t i;
for (i = 0; i != 3; i++) {
fprintf(stdout, "%zu: %s\n", i, s[i]);
}
return 0;
}
Note that the original declaration
char* s[4][10] = { "abc\0", "aab\0", "cbb\0" };
had several errors:
char * instead of char
[4][10] instead of [10][4]
"abc\0" instead of "abc" (the trailing '\0' is implicit).