Thread: explanation

for this program, I prompt the use to enter a string. after the string is executed, the program will then execute to convert all lowercase letters into upper case.

Code:

#include<stdio.h>
#include <ctype.h>
#include <string.h>

main()
{
	char x[100];
	int i = 0;

	printf("Enter a string: ");	
	gets(x);

	for(i = 0; x[i] !=0; i++)
	{
		x[i] = toupper(x[i]);
	}

	printf("%s\n", x);
}

But when I use another char array to represent the new uppercased strings it would output an extra character.

Code:

#include<stdio.h>
#include <ctype.h>
#include <string.h>

main()
{
	char x[100];
	char l[100];
	int i = 0;

	printf("Enter a string: ");	
	gets(x);

	for(i = 0; x[i] !=0; i++)
	{
		l[i] = toupper(x[i]);
	}

	printf("%s\n", l);
}

Sample run for the first program.

Code:

Enter a string: This is A teSt
THIS IS A TEST

Sample run for the 2nd program with an l char array:

Code:

Enter a string: This is A teSt
THIS IS A TEST9

As you can see, there's an excess char '9' in the executed program. Can anyone explain as to why this happens?

There is no NULL terminator character in the second array to tell printf "here is where the string stops". You will need to add the terminator, or initialize your arrays with 0s

Code:

char l[100] = { 0 };

Also, you should not be using gets. gets is BAD.

thank you!