Programming Forums > Application Development > C

Uday · Mar 29th, 2006, 7:23 AM

This program is not terminating. Can anyone tell me what is the problem with the code.

#include<Stdio.h>
#include<String.h>
void main()
{
int i=0;
char str1[10];
clrscr();
printf("Enter string1");
scanf("%s",str1);
for(i=0;str1!='\0';i++)
{
printf("%d",i);
}
getch();
}

Is that all strings end with the null character '\0'

nnxion · Mar 29th, 2006, 7:44 AM

I sure hope you are kidding...
What in the world are you trying to do?

(Toggle Plain Text)

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

int main()
{
	int i;
	char str1[10];
	printf("Enter string1\n");
	scanf("%s",str1);
	printf("%s", str1);
	getch();
	return 0;
}

Yes C strings end with the null character. You don't have to tell us what it is, believe me, we know.

The Dark · Mar 29th, 2006, 7:54 AM

(Toggle Plain Text)

for(i=0;str1!='\0';i++)
{
printf("%d",i);
}

Your loop condition is on str1!='\0', but you don't change the value of str1 inside the loop. I suspect you meant

(Toggle Plain Text)

for(i=0;str1[i]!='\0';i++)

Uday · Mar 29th, 2006, 9:33 PM

I wrote this program to test whether the sting ends with '\0'. I noticed that there is some problem with this part (I was trying another program that is a bigger one. That program is posted below). I corrected the above one. I had to put str1[i]

The original program that is not working is

Q)
Write a program to transform its input according to a specified transformation scheme. The transformation scheme will consist of two strings: a string of characters and then a string of replacement characters. The idea is that your program replaces every instance of the i'th character in the initial string with the i'th character in the replacement string (see example output below). Your transformation scheme can include letters (uppercase and lowercase), numbers and punctuation. When no substitution is defined for a character, the program just passes it through to the output unchanged. Your program should check that each transformation scheme contains two strings and that the strings contain the same number of characters. The program should inform the user of any errors in the transformation scheme. Your program should also have a list of 10 phrases to which it will apply transformation schemes. For each of these phrases, your program writes the phrase before and after the substitutions have been made.

Example:

Input Line: abcde bcead

Output:
Transformation Scheme - abcde bcead

alex was here
bldx wbs hdrd

a better example
b cdttdr dxbmpld

(Toggle Plain Text)

#include<Stdio.h>
#include<String.h>
void main()
{
int i,j;
char phrase[10][20],charstr[10],repstr[10];
clrscr();
printf("Enter 10 phrases to be tranformed");
 for(i=0;i<10;i++)
  {
   scanf("%s",phrase[i]);
  }
printf("Enter the character string\n");
scanf("%s",charstr);
printf("Enter the replacement string\n");
scanf("%s",repstr);
i=0;
while(repstr[i]=='\0')
 {
  printf("The Repstr is null and the output is\n");
  for(i=0;i<10;i++)
   {
    printf("%s\n",phrase[i]);
   }
 }
i=strlen(repstr);
j=strlen(charstr);
printf("%d%d",i,j);
if(i==j)
 {
   for(i=0;i<10;i++)
   printf("%s",phrase[i]);
   for(j=0;phrase[i][j]!='\0';j++)
	{
	if(phrase[i][j]==charstr[j])
		{
		phrase[i][j]=repstr[j];
	 printf("%s",phrase[i]);
		}
	}
 }
else
 {
printf("The transformation cannot take place since charstr!=repstr");
 }
getch();
}

DaWei · Mar 29th, 2006, 10:23 PM

I'm going to suggest that you go through your loops and other machinations once with a pencil and paper, tracking the array index and pointer values of anything you use. Make sure you DO understand C-string char arrays and pointers. You'll pick up a lot that will stick in your mind, that way. I would suggest that you limit the number of input characters allowed to prevent buffer overflow and the attendant crashes and other headaches. You may do that with scanf, or you may switch to fgets. Consult your documentation.

The Dark · Mar 29th, 2006, 11:10 PM

Your welcome, by the way.

Mar 29th, 2006, 7:23 AM	#1
Uday Programmer Join Date: Oct 2005 Location: India Posts: 30 Rep Power: 0	Problem with strings This program is not terminating. Can anyone tell me what is the problem with the code. (Toggle Plain Text) #include<Stdio.h> #include<String.h> void main() { int i=0; char str1[10]; clrscr(); printf("Enter string1"); scanf("%s",str1); for(i=0;str1!='\0';i++) { printf("%d",i); } getch(); } #include<Stdio.h> #include<String.h> void main() { int i=0; char str1[10]; clrscr(); printf("Enter string1"); scanf("%s",str1); for(i=0;str1!='\0';i++) { printf("%d",i); } getch(); } Is that all strings end with the null character '\0'

Mar 29th, 2006, 7:44 AM	#2
nnxion Programming Guru Join Date: Jun 2005 Location: elemental plane Posts: 1,429 Rep Power: 5	I sure hope you are kidding... What in the world are you trying to do? (Toggle Plain Text) #include<stdio.h> #include<string.h> #include <conio.h> int main() { int i; char str1[10]; printf("Enter string1\n"); scanf("%s",str1); printf("%s", str1); getch(); return 0; } #include<stdio.h> #include<string.h> #include <conio.h> int main() { int i; char str1[10]; printf("Enter string1\n"); scanf("%s",str1); printf("%s", str1); getch(); return 0; } Yes C strings end with the null character. You don't have to tell us what it is, believe me, we know. __________________ "Employ your time in improving yourself by other men's writings, so that you shall gain easily what others have labored hard for." -- Socrates

Mar 29th, 2006, 7:54 AM	#3
The Dark Expert Programmer Join Date: Jun 2005 Posts: 894 Rep Power: 4	(Toggle Plain Text) for(i=0;str1!='\0';i++) { printf("%d",i); } for(i=0;str1!='\0';i++) { printf("%d",i); } Your loop condition is on str1!='\0', but you don't change the value of str1 inside the loop. I suspect you meant (Toggle Plain Text) for(i=0;str1[i]!='\0';i++) for(i=0;str1[i]!='\0';i++)

Mar 29th, 2006, 9:33 PM	#4
Uday Programmer Join Date: Oct 2005 Location: India Posts: 30 Rep Power: 0	I wrote this program to test whether the sting ends with '\0'. I noticed that there is some problem with this part (I was trying another program that is a bigger one. That program is posted below). I corrected the above one. I had to put str1[i] The original program that is not working is Q) Write a program to transform its input according to a specified transformation scheme. The transformation scheme will consist of two strings: a string of characters and then a string of replacement characters. The idea is that your program replaces every instance of the i'th character in the initial string with the i'th character in the replacement string (see example output below). Your transformation scheme can include letters (uppercase and lowercase), numbers and punctuation. When no substitution is defined for a character, the program just passes it through to the output unchanged. Your program should check that each transformation scheme contains two strings and that the strings contain the same number of characters. The program should inform the user of any errors in the transformation scheme. Your program should also have a list of 10 phrases to which it will apply transformation schemes. For each of these phrases, your program writes the phrase before and after the substitutions have been made. Example: Input Line: abcde bcead Output: Transformation Scheme - abcde bcead alex was here bldx wbs hdrd a better example b cdttdr dxbmpld (Toggle Plain Text) #include<Stdio.h> #include<String.h> void main() { int i,j; char phrase[10][20],charstr[10],repstr[10]; clrscr(); printf("Enter 10 phrases to be tranformed"); for(i=0;i<10;i++) { scanf("%s",phrase[i]); } printf("Enter the character string\n"); scanf("%s",charstr); printf("Enter the replacement string\n"); scanf("%s",repstr); i=0; while(repstr[i]=='\0') { printf("The Repstr is null and the output is\n"); for(i=0;i<10;i++) { printf("%s\n",phrase[i]); } } i=strlen(repstr); j=strlen(charstr); printf("%d%d",i,j); if(i==j) { for(i=0;i<10;i++) printf("%s",phrase[i]); for(j=0;phrase[i][j]!='\0';j++) { if(phrase[i][j]==charstr[j]) { phrase[i][j]=repstr[j]; printf("%s",phrase[i]); } } } else { printf("The transformation cannot take place since charstr!=repstr"); } getch(); } #include<Stdio.h> #include<String.h> void main() { int i,j; char phrase[10][20],charstr[10],repstr[10]; clrscr(); printf("Enter 10 phrases to be tranformed"); for(i=0;i<10;i++) { scanf("%s",phrase[i]); } printf("Enter the character string\n"); scanf("%s",charstr); printf("Enter the replacement string\n"); scanf("%s",repstr); i=0; while(repstr[i]=='\0') { printf("The Repstr is null and the output is\n"); for(i=0;i<10;i++) { printf("%s\n",phrase[i]); } } i=strlen(repstr); j=strlen(charstr); printf("%d%d",i,j); if(i==j) { for(i=0;i<10;i++) printf("%s",phrase[i]); for(j=0;phrase[i][j]!='\0';j++) { if(phrase[i][j]==charstr[j]) { phrase[i][j]=repstr[j]; printf("%s",phrase[i]); } } } else { printf("The transformation cannot take place since charstr!=repstr"); } getch(); }

Mar 29th, 2006, 10:23 PM	#5
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	I'm going to suggest that you go through your loops and other machinations once with a pencil and paper, tracking the array index and pointer values of anything you use. Make sure you DO understand C-string char arrays and pointers. You'll pick up a lot that will stick in your mind, that way. I would suggest that you limit the number of input characters allowed to prevent buffer overflow and the attendant crashes and other headaches. You may do that with scanf, or you may switch to fgets. Consult your documentation. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Mar 29th, 2006, 11:10 PM	#6
The Dark Expert Programmer Join Date: Jun 2005 Posts: 894 Rep Power: 4	Your welcome, by the way.

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Programming Forums > Application Development > C

Problem with strings