Newcomer to C

[spinC]

Hi, I am somewhat of a new programmer and the only experience I have was using Java. But anways, I am now learning to write a C program and am unsure of some of the new syntaxs and restrictions. I looked at a sample program that prints out the alphabet from A-Z.
I am trying to understand this code, and when I rewrite it in my own words it works.
One thing I am unsure of is the structure of my loop(s), I can get it to print out all the characters but I am trying to make the letters of the alphabet take up 3 lines rather than all on 1 line. Any suggestions are gratefully appreciated.
Thanks in advance.

[PaulWendt]

I don't really understand what your question is. If you're asking
how to go to the next line, you use the newline character: '\n'.
With this code:

#include <stdio.h>
int main(int argc, char* argv[])
{
   printf("line1\n");
   printf("this is on line2\n");
   printf("this is on line 3\n");
   return 0;
}

The following will be output:

line1
this is on line2
this is on line 3

--Paul

[spinC]

Thanks for the reply. That kinda helped,
but my output is getting printed out with a For Loop
because my text is in an array? Any ideas..?
Thanks again

[Eric Leung]

Hi spinC ,
Is this what you want?

for(int i=0;i<ArrayCount;i++)
{
printf("%s\n",Array[i]);
}

[spinC]

That code printed out the same element of my array 2 times on the line. Not exactly..... I'll post my code to show what I have and what I am trying to do.

int linenum=0;
int count=0;
char alpha[26] = "abcdefghijklmnopqrstuvxyz";

for(i = 0;i <= strlen(alpha); i++)
{
printf("%c\n", alpha[i]);
}

return 0;

This code prints out every letter on an individual line.
I am trying to have a variable where I can print out the entire
alphabet but I can specify how many lines to print it out on.

ie. abcdef
ghijklm
nopqrs
tuvwxy
z

5 lines

[Eric Leung]

Hi spinC,

I am trying to understand what you want, but your ie. puzzled me.

ie. abcdef //6 characters
ghijklm //7 characters ??
nopqrs //6 characters
tuvwxy //6 characters
z //1 character

If you want to print out 6 characters one line,hope this helps,

int linenumber=0;
int count=0;
char alpha[26] = "abcdefghijklmnopqrstuvxyz";
for(int i=0;i<strlen(alpha);i++)
{
printf("%c",alpha[i]);
count++;
if(count==6)
{
printf("\n");
linenumber++;
count=0;
}
}
printf("\n\n");
if(count!=0)
{
linenumber++;
}
printf("%d lines\n",linenumber);

[Kheun]

This code example should print out the array in the specific no. of lines.

		char alpha[]="abcdefghijklmnopqrstuvwxyz";

	const int lines = 5;
	const int len = strlen(alpha);
	
	// Equivalent to len/lines and then round-up.
	int charPerLine = (len + lines - 1)/lines; 

	for(int i=0; i<len; ++i)
	{
		if(i && (i%charPerLine) == 0)
			printf("\n");

		printf("%c", alpha[i]);
	}

BTW, there is a mistake in the previous example,
char alpha[26] = "abcdefghijklmnopqrstuvxyz";

You need to allocate an extra space for '\0'.

[spinC]

hmmmm.....I think what you posted might help but Im basically just trying to get it to print out the alphabet with a set number of characters per line, and this set number of characters per line is a number input by the user. Sorry I made a typo in the example so I see how it could be confusing. Thanks again for all the help though.

ie. if they enter the value of 5

abcde
fghij
klmno
pqrst
uvwxy
z

[dav79]

try it:

void f(int s){
	int i;
	char alpha[]="abcdefghijklmnopqrstuvwxyz";
	
	for(i = 0;i <= strlen(alpha); i++)
	{
		if(!(i%s)) printf("\n");
		printf("%c", alpha[i]);
	}
}

void main(){
	f(5);
return;
}

[Paul McKenzie]

Originally posted by dav79
try it:

void main(){
}

If they are using an ANSI C++ compiler, this won't compile. The main() function returns an int, not void.

Regards,

Paul McKenzie

[spinC]

thank you all for the replies and help....
its good to go and prints out fine.
thanks again