Outputting the size of an array

[tigrfire]

I need the following code to return a string of character stored in
array and also return the size of the array. I want to do it without
using pointers and so far have the character storing working fine, it's
just that I can't really think of a way to return the number of
characters in the array. Anyone help me out?

int inputPhrase()
{
char msg[characters], ch;
int i = 0;
int n = 5;

while ((ch = getchar()) != '\n')
{
msg[i++] = ch;
}

msg[i] = '\0'; /* prevents garbage from being place at end of array
output */

i = 0;

while (msg[i] != '\0')
{
printf("%c", msg[i++]);
}

return n;
}

int main()
{
inputPhrase();

int size = inputPhrase();
printf("%d\n", size);

return 0;
}

[tigrfire]

Ignore the n = 5; I just used that to make sure that main was
outputting n correctly, obviously, n needs to be originally set to 0
then incremented up to the appropriate number of characters in the
array.

[Sandeep]

tigrfire wrote:

I need the following code to return a string of character stored in
array and also return the size of the array. I want to do it without
using pointers and so far have the character storing working fine, it's
just that I can't really think of a way to return the number of
characters in the array. Anyone help me out?

int inputPhrase()
{
char msg[characters], ch;
int i = 0;
int n = 5;

while ((ch = getchar()) != '\n')
{
msg[i++] = ch;
}

msg[i] = '\0'; /* prevents garbage from being place at end of array
output */

i = 0;

while (msg[i] != '\0')
{
printf("%c", msg[i++]);
}

return n;
}

int main()
{
inputPhrase();

int size = inputPhrase();
printf("%d\n", size);

return 0;
}

You have initialized n, but never used it !

while ((ch = getchar()) != '\n')
{
msg[i++] = ch;
}

n = i; // This should set the value of n .

Further ,

char msg[characters],

This is not a valid statement, an array cannot be initialized with a
variable.

[tigrfire]

Outside the function I defined characters as:
#define characters 30

If I do as you say and modify my code so that n = i on the line you
reference to, the program's output is this:

this is sample output //entered string of chars
this is sample output //returned string of chars
0 //the value of n

[Sandeep]

tigrfire wrote:

Outside the function I defined characters as:
#define characters 30

If I do as you say and modify my code so that n = i on the line you
reference to, the program's output is this:

this is sample output //entered string of chars
this is sample output //returned string of chars
0 //the value of n

int main()
{
inputPhrase();

int size = inputPhrase();
printf("%d\n", size);

return 0;
}

You are calling "inputPhrase(); " twice ! . remove it from the line
after main(){. It works fine for me.

[haroon]

You are not setting `n' to the number of characters input, this might
help.

/*CODE BEGINS*/

int inputPhrase()
{
char msg[characters], ch;
int i = 0;
int n = 0;
while ((ch = getchar()) != '\n')
{
msg[i++] = ch;
}
msg[i] = '\0'; /* prevents garbage from being place at end of array
output */
/*****/
n = i;
/*****/
i = 0;
while (msg[i] != '\0')
{
printf("%c", msg[i++]);
}
return n;
}

int main()
{
int size = inputPhrase();
printf("%d\n", size);
return 0;
}
/*CODE ENDS*/

and remove that extra call to inputPhrase from your main.

HTH.

[Jack Klein]

On 14 Nov 2005 18:34:00 -0800, "tigrfire" <bb******@gmail .com> wrote
in comp.lang.c:

Outside the function I defined characters as:
#define characters 30

Then you should follow the very commonly used convention and define
macros in ALL UPPER CASE.

#define CHARACTERS 30

That way, you would have posted:

char msg[CHARACTERS], msg;

....and most people would have assumed it was a macro.

--
Jack Klein
Home: http://JK-Technology.Com
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http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html

[Anand]

Sandeep wrote:

tigrfire wrote:
[...]
Further ,

char msg[characters],

This is not a valid statement, an array cannot be initialized with a
variable.

That's not true, C99 has VLAs (Variable Length Arrays) where in you can
use a variable to initialize the array size.

[Sandeep]

That's not true, C99 has VLAs (Variable Length Arrays) where in you can
use a variable to initialize the array size.

I stand corrected. I tried it and it is working.

[tigrfire]

I've modified my code as you mentioned and also made some syntatical
changes that Jack recommended. I'm now trying to do a number of things
with my code:
1. I'm writing a function called cleanUpPhrase that removes
punctuation and whitespace. I'm going to try and accomplish this using
tolower() and isAlpha() from the ctype library I've included. I'm just
a little confused on how I can pass the string of chars from my
inputPhrase function into my cleanUpPhrase function.
2. On another note, I'm also trying to modify my inputPhrase
function so that if the user-inputted string of chars is longer than 30
chars long, it will truncate the message and only output the first 30
chars. In my main function I used an if loop to truncate the size, but
nothing I've tried in my inputPhrase function to truncate the actual
returned message output is working.

Here's the code:
// code starts

#include <stdio.h> /* provides scanf, printf, getchar */
#include <ctype.h> /* provides tolower, isalpha */

#define MAX_CHARS 30 /* defines the value of characters equal to 30 as
a global
constant */

int inputPhrase(); /* This function reads in a phrase, maximum of 30
characters,
entered by the user. Function returns the phrase, stored in
a character array. Also returns the size of the array. Only
the first 30 characters are stored in the array if the array
size is greater than 30. */

int inputPhrase()
{
char msg[MAX_CHARS], ch;
int i = 0;
int n = 0;
while((ch = getchar()) != '\n')
{
msg[i++] = ch;
}
msg[i] = '\0'; /* prevents garbage from being place at end of array
output */

n = i;

i = 0;

while(msg[i] != '\0')
{
printf("%c", msg[i++]);
}
return n;

}

void cleanUpPhrase()
{

}

int main()
{
int size = inputPhrase();

if(size >= MAX_CHARS)
{
size = 30;
}

printf("\n%d\n" , size);
return 0;
}

//code ends