Infinite loop in C counting spaces - c

Question

I'm writing a program to count spaces and vowels and it didn´t work, I think I did an infinite loop.I'll show you my code:
int contar_p(char a[100]) {
int i = 0, spaces = 1;
while (a[i] != '\0' && i < 100) {
if (a[i] == ' ') {
spaces += 1;
i++;
}
}
return spaces;
}
int contar_v(char b[100]) {
int i = 0, counter = 0;
while (b[i] != '\0' && i < 100) {
if (b[i] == 'a' || b[i] == 'e' || b[i] == 'i' || b[i] == 'o' || b[i] == 'u') {
counter += 1;
}
i++;
}
return counter;
}
int main(void){
char phrase[100];
int words = 0, vowels = 0;
printf("write a phrase ");
gets(phrase);
palabras = contar_p(phrase);
vocales = contar_v(phrase);
printf("%d\n", words);
printf("%d", vowels);
return 0;
}

Answer 1

The loop
while (a[i]!='\0'&&i<100){
if(a[i]==' '){
spaces+=1;
i++;
}
}
is an infinite loop. Place i++ outside the if. Change it to
while (a[i]!='\0'){ // No need of condition i < 100
if(a[i]==' '){
spaces+=1;
}
i++;
}

Answer 2

Maybe another approach will help you to understand things easier, i'm mean you do know that there are A,E,I,O,U also and not only a,e,i,o,u. You should never use gets instead use fgets, anyway take a look at the following program:
#include <stdlib.h>
#include <stdio.h>
void countVowels(char* array){
int i,j,v;
i=0;
int count = 0;
char vowel[]={'a','e','i','o','u','A','E','I','O','U'};
while(array[i]!='\0'){
for(v=0;v<10;v++){
if (array[i]==vowel[v]){
j=i;
while(array[j]!='\0'){
array[j]=array[j+1];
j++;
}
count++;
i--;
break;
}
}
i++;
}
printf("Found %d Vowels\n",count);
}
void contar_p(char a[100]) {
int i = 0, spaces = 0;
for(i=0;a[i]!='\0';i++){
if(a[i]==' ')
spaces++;
}
printf("Found %d Spaces\n",spaces);
}
int main(void){
char a[]="aa bb EOU cc ii";
countVowels(a);
contar_p(a);
return 0;
}
Output:
Found 7 Vowels
Found 4 Spaces