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I need to write a procedure that takes a pointer to a string p and shows(prints) the number of occurrences of each letter within the word. Example: macaw should show a=3 and r=2 shoe should show s=1, a=2, t=1 and o=1
I was trying to think of a logic in pure C but I can’t. Could you help me?
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void verifiRepetition(char [],int);
void check Existence(char [],char, int *);
int main(){
int const quantodadeElementos = 50;
char vetorPalavra[quantodadeElementos];
printf("Informe uma palavra: ");
scanf("%s", vetorPalavra);
verificarRepeticao(vetorPalavra,quantodadeElementos);
return 0;
}
void checkRepetition(char vectorPalavra[], int quantityElements){
int contador = 0, i = 0, j = 0, teste = 0;
char charVerificaAtual, vetorPalavraDois[quantidadeElementos];
for(i=0;i<quantidadeElementos;i++){
vetorPalavraDois[i] = '\0';
}
for(i=0;vetorPalavra[i]!='\0';i++){
charVerificaAtual = vetorPalavra[i];
verificarExistencia(vetorPalavraDois,charVerificaAtual,&teste);
if(teste == 1){
}else if(teste == 0){
for(j=0;vetorPalavra[j]!='\0';j++){
if(charVerificaAtual == vetorPalavra[j]){
contador++;
}
}
vetorPalavraDois[i]=charVerificaAtual;
printf("%c Ocorreu %d.\n", charVerificaAtual,contador);
contador = 0;
teste = 0;
}
}
} void checkExistence(char vectorPalavraDois[], char charVerificaAtual, int *test){
int i=0, testedois = 0;
for(i=0;vetorPalavraDois[i] != '\0';i++){
if(charVerificaAtual == vetorPalavraDois[i]){
testedois = 1;
break;
}
}
*teste = testedois;
}
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void procedure (char *palavra){
int cont, x=0, cont2;
char letra;
for (cont=0; cont<strlen(palavra); cont++){
letra=palavra[cont];
for (cont2=0; cont2<strlen(palavra); cont2++){
if(letra==palavra[cont2]){
x++;
}
}
printf ("\nocorrencias da letra %c = %d", letra, x);
x=0;
}
}
main(){
char word[20];
char *p = word;
printf ("Digite uma palavra: ");
fflush(stdin);
gets(p);
procedure(p);
}
Avoid answering only with code. Explain your solution a little bit.
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Set an array of integers of sufficient size for each letter of the alphabet (letter[27]) and initialize with 0. Scroll through its string, character by character, and taking advantage of a particularity of C do: letter[tplower(word[i]) - 'a']+++. At the end print each letter element that is non-zero. To know which letter is just add 'a' to the index.
– anonimo