Reputation:
I'm trying to creat a program that removes all occurences from a character that I choose from a certain string and also returs the number of characters that were removed.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define DIMV 10
int eliminar(char texto[], char ch, char novoTexto[]){
int a=0,i;
for(i=0; texto[i] != '\0';i++){
texto[i] = texto[i+a];
novoTexto[i]=texto[i];
if(novoTexto[i]==ch){
novoTexto[i] = '\0';
a++;
}
}
return a;
}
int main()
{
char frase[]="Uma arara torta!";
char res[50];
printf("%s\n",frase);
printf("%d\n",eliminar(frase,'r',res));
printf("%s\n",res);
return 0;
}
When I run the program, it returns:
Uma arara torta!
3
Uma a
What I wanted to return is:
Uma arara torta!
3
Uma aaa tota!
It's a silly question but I can't find the mistake that I made. Thank you!
Upvotes: 0
Views: 881
Reputation: 11
Putting '\0' means to terminate the string there. check this:
int eliminator(char * texto, char ch, char * novoTexto) {
int i=0;
while(*texto != '\0') {
if (*texto == ch) {
texto++;
++i;
} else {
*novoTexto++ = *texto++;
}
}
*novoTexto = '\0';
return i;
}
int main(int arg, char **argv) {
char frase[] = "Uma arara torta!";
char res[50];
printf("%s\n",frase);
printf("%d\n",eliminator(frase, 'r', res));
printf("%s\n",res);
}
Upvotes: 0
Reputation: 75062
Putting '\0'
means to terminate the string there.
You have to skip adding characters to eliminate.
Try this:
int eliminar(char texto[], char ch, char novoTexto[]){
int a=0,i,j=0;
for(i=0; texto[i] != '\0';i++){
if(texto[i]==ch){
a++;
}else{
novoTexto[j]=texto[i];
j++;
}
}
novoTexto[j]='\0';
return a;
}
Upvotes: 3