Homework help... Again.

[rafaelcbf]

So i'm wirghting this program:

#include <stdio.h>
#include <iostream>

int main()
{
char A,a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z;
printf ("Introduzca un caracter, por favor.");
scanf("%c", &A);
if (A==b,c,d,f,g,h,j,k,l,m,n,p,q,r,s,t,v,w,x,y,z)
printf("Caracter es una consonante.");
else (A==a,e,i,o,u)
printf ("Caracter es una vocal.");
else (A==%i)
printf ("Caracter es un numero.");
} 

It's supposed to:

1.-Ask you to input a (1) charecter/intager.
2.- Then it should output wether if it's a consonant, vowel or integer.

That's it, but I'm still having loads of trouble with characters, I can I get some orientation? Please.

[Cookieh]

I have made a code in C which looks a bit like this

#include<stdio.h> #include<conio.h> #include<string.h> void main() {  char s[100];  int b,c,d,e,i,l,p;  b=c=d=e=0;  clrscr();  printf("\n\t Enter the string :=>");  gets(s);  l=strlen(s);  for(i=0;s[i]!='\0';i++)   {    if((s[i]>=65&&s[i]<=90||s[i]>=97&&s[i]<=122))    b++;    if(s[i]>=48&&s[i]<=57)    c++;    if(s[i]=='a'||s[i]=='e'||s[i]=='i'||s[i]=='o'||s[i]=='u'||s[i]=='A'||s[i]=='E'||s[i]=='I'||s[i]=='O'||s[i]=='U')    d++;       if (s[i]==' ')    e++;    }    p=b-d;    printf("\n\t No of vowels=%d \n",d);    printf("\n\t No of consonants=%d \n ",p);    printf("\n\t No of digits=%d \n",c);    printf("\n\t no of words=%d \n ",e+1);    printf("\n\t No of specal character=%d \n",l-b-c-e);
getch();
}

The only thing is that this is not my complete version as my development version is on my mac which is currently not usable. I think that this would be pretty simple to recreate in php.. Good Luck

[Cookieh]

Sorry, I can not get it to be in the formating which has indentations, hopefully you can solve it... Sorry

[rafaelcbf]

#include<stdio.h>
 #include<conio.h>
 #include<string.h>
 void main()
 {
  char s[100];
  int b,c,d,e,i,l,p;
  b=c=d=e=0;
  clrscr();
  printf("\n\t Enter the string :=>");
  gets(s);
  l=strlen(s);
  for(i=0;s[i]!='\0';i++)   {    if((s[i]>=65&&s[i]<=90||s[i]>=97&&s[i]<=122))    b++;
    if(s[i]>=48&&s[i]<=57)    c++;    if(s[i]=='a'||s[i]=='e'||s[i]=='i'||s[i]=='o'||s[i]=='u'||s[i]=='A'||s[i]=='E'||s[i]=='I'||s[i]=='O'||s[i]=='U')    d++;
       if (s[i]==' ')    e++;    }
    p=b-d;
    printf("\n\t No of vowels=%d \n",d);
    printf("\n\t No of consonants=%d \n ",p);
    printf("\n\t No of digits=%d \n",c);
    printf("\n\t no of words=%d \n ",e+1); 
   printf("\n\t No of specal character=%d \n",l-b-c-e);
getch();
} 

I see what it does, but this is for a whole string, I just need it to be for one (1) char or int. I guess I could kind of re work it. Thank you. Giving me the answer is not the same as help though, can anyone tell me more or less what I'm doing wrong? Please.

[Arndt]

So i'm wirghting this program:

#include <stdio.h>
#include <iostream>

int main()
{
char A,a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z;
printf ("Introduzca un caracter, por favor.");
scanf("%c", &A);
if (A==b,c,d,f,g,h,j,k,l,m,n,p,q,r,s,t,v,w,x,y,z)
printf("Caracter es una consonante.");
else (A==a,e,i,o,u)
printf ("Caracter es una vocal.");
else (A==%i)
printf ("Caracter es un numero.");
} 

It's supposed to:

1.-Ask you to input a (1) charecter/intager.
2.- Then it should output wether if it's a consonant, vowel or integer.

That's it, but I'm still having loads of trouble with characters, I can I get some orientation? Please.

I recommend that you read, refresh your memory or ask your teacher about these things:

1) character constants

2) tests and logical operators

Your program seems to be intended to be C++, judging from the include files, but you should say so explicitly - many languages are discussed here.

[rafaelcbf]

Oh, sorry about that, yes this is c++.

[lisati]

i recommend that you read, refresh your memory or ask your teacher about these things:

1) character constants

2) tests and logical operators

Agreed
The way the first listing reads to me is as if it is comparing variables, which is not necessarily the same thing as testing the variables against predefined values.

If you do something like:

if (A=b)....

A and b are labels that refer to memory locations that could contain any information.

[Tony Flury]

Oh, sorry about that, yes this is c++.

If it is C++ then you should not really use stdio (that is C), and you should use cin and cout (with the >> and <<) operators for input and output).

[Arndt]

If it is C++ then you should not really use stdio (that is C), and you should use cin and cout (with the >> and <<) operators for input and output).

But all he needs to do is remove "#include <iostream>" and it becomes C.

[Tony Flury]

Arndt - very true - he needs to decide one way or another - or taught one way or another