Programming Forums > Application Development > C++

TecBrain · Apr 13th, 2005, 12:12 AM

If the user input a space the output should be replaced with a ' * ', Not seems to except, I have added the value 32 which represent the space value on the ASCII table,However; if I replace the value of 32 to 33 which represent ' ! ' char, it would work, Any idea!!

(Toggle Plain Text)

void strencypt(char str[])
{
	char star='*';
	char str2[10];
	int i;
	
	for(i=0;((i<10) && (str[i]!='\0'));i++)
	{
			if ((int)(str[i])==32)
			str2[i]= star;
	
			else str2[i]=(int)(str[i]+1);

	}
	
	std::cout << "String A Is Encrypted Now With This Format: " << "\n";
	
	for(i=0;((i<10) && (str[i]!='\0'));i++)
	{
		std::cout << str2[i];
	}
	std::cout << "\n";
	
}

kirkl_uk · Apr 13th, 2005, 3:02 AM

Hoping I understand this right, why don't you loop through the string, using strlen for the counter termination, and then using something simple like "if(str[i] == ' ') str2[i] = '*';" ? I did something similar to this in a previous project where I replaced a space with a _ for writing/reading files.

kirkl_uk

TecBrain · Apr 13th, 2005, 3:16 AM

I have tried that before and I ended up with the same results, The loop never stops .....

(Toggle Plain Text)

void strencypt(char str[])
{
	char star='*';
	char str2[10];
	int i;
	
	for(i=0;((i<10) && (str[i]!='\0'));i++)
	{
		if((str[i])== ' ') str2[i]= star;
	
		else str2[i]=int(str[i]+1);

	}
	
	std::cout << "String A Is Encrypted Now With This Format: " << "\n";
	
	for(i=0;((i<10) && (str[i]!='\0'));i++)
	{
		std::cout << str2[i];
	}
	std::cout << "\n";
	
}

kirkl_uk · Apr 13th, 2005, 3:43 AM

Why do you limit str2 to 10 chars? I think I need more info to help on this one.

MSN me if you like
kirkl_uk@hotmail.com

TecBrain · Apr 13th, 2005, 4:39 AM

The Problem is not the size of the array!!!! its to replace the space with a ' * '!!!!!

kirkl_uk · Apr 13th, 2005, 4:49 AM

I am aware of this. However, your explanation of the problem is very poor. Note how no one is able to help.

TecBrain · Apr 13th, 2005, 6:35 AM

Is it!!! I dont see any hard part on it, However;

The problem with this code is the loop never stops when it reaches the space character.

am not sure if the problem is my compiler

This is the output of entering the string "hell o"

(Toggle Plain Text)

EXECUTING:
/home/moderator/C++/project2/main/readstri
----------------------------------------------
Enter 1 : To Read In A String
Enter 2 : To Find The Size Of The String
Enter 3 : To Copy String A into Sting B
Enter 4 : To Merger String A With Sting B
Enter 5 : To Compare Two Strings
Enter 6 : To Encrypt String A, Method A
Enter 7 : To Encrypt String A, Method B
Enter 8 : To Locate a Character
Enter 9 : To Exit The Program
6

Enter A String
hell o

String A Is Encrypted Now With This Format:
ifmm

Enter 1 : To Read In A String
Enter 2 : To Find The Size Of The String
Enter 3 : To Copy String A into Sting B
Enter 4 : To Merger String A With Sting B
Enter 5 : To Compare Two Strings
Enter 6 : To Encrypt String A, Method A
Enter 7 : To Encrypt String A, Method B
Enter 8 : To Locate a Character
Enter 9 : To Exit The Program
6

Enter A String
hell o

String A Is Encrypted Now With This Format:
ifmm

Enter 1 : To Read In A String
Enter 2 : To Find The Size Of The String
Enter 3 : To Copy String A into Sting B
Enter 4 : To Merger String A With Sting B
Enter 5 : To Compare Two Strings
Enter 6 : To Encrypt String A, Method A
Enter 7 : To Encrypt String A, Method B
Enter 8 : To Locate a Character
Enter 9 : To Exit The Program
6

Enter A String
hell o

String A Is Encrypted Now With This Format:
ifmm

Enter 1 : To Read In A String
Enter 2 : To Find The Size Of The String
Enter 3 : To Copy String A into Sting B
Enter 4 : To Merger String A With Sting B
Enter 5 : To Compare Two Strings
Enter 6 : To Encrypt String A, Method A
Enter 7 : To Encrypt String A, Method B
Enter 8 : To Locate a Character
Enter 9 : To Exit The Program
6

Enter A String
hell o

String A Is Encrypted Now With This Format:
ifmm
.
.
.
.
.
.

The loop wont stop running, if you complie it you will see what I mean

(Toggle Plain Text)

void strencypt(char str[])
{
	char star='*';
	char str2[10];
	int i;
	
	for(i=0;((i<10) && (str[i]!='\0'));i++)
	{
		if((str[i])== ' ') str2[i]= star;
	
		else str2[i]=int(str[i]+1);

	}
	
	std::cout << "String A Is Encrypted Now With This Format: " << "\n";
	
	for(i=0;((i<10) && (str[i]!='\0'));i++)
	{
		std::cout << str2[i];
	}
	std::cout << "\n";
	
}

and this is the part is on my main body

(Toggle Plain Text)

		case 6:
		{
			std::cout << "Enter A String" << "\n";
			std::cin >> name;
			strencypt(name);
			break;

kirkl_uk · Apr 13th, 2005, 7:05 AM

We are here to help, so try not to get so defensive. It's clear to you how the program should work and what happens during execution, but it wasn't clear for an outsider to the program. I don't have a compiler on this machine at the moment, so I will have a look at this when I get home later tonight. Remember, I am trying to help you.

TecBrain · Apr 13th, 2005, 8:04 AM

kirkl_UK I wasn't trying to offend you, however; if I did please accept my apology.

some how I think I need to use the command gets instead of cin!!!Not sure though I'll try it.

spydoor · Apr 13th, 2005, 9:14 AM

wierd.. you thought of the answer but don't know why.

(Toggle Plain Text)

cin >> name;  // reads up to whitespace, newline , or EOF

so you were never reading in a space to begin with.

(Toggle Plain Text)

cin.getline(name, length); // reads up to newline, or EOF

Apr 13th, 2005, 12:12 AM	#1
TecBrain Hobbyist Programmer Join Date: Sep 2004 Location: Cyprus Posts: 147 Rep Power: 4	*replace space with ' '** If the user input a space the output should be replaced with a ' * ', Not seems to except, I have added the value 32 which represent the space value on the ASCII table,However; if I replace the value of 32 to 33 which represent ' ! ' char, it would work, Any idea!! (Toggle Plain Text) void strencypt(char str[]) { char star=''; char str2[10]; int i; for(i=0;((i<10) && (str[i]!='\0'));i++) { if ((int)(str[i])==32) str2[i]= star; else str2[i]=(int)(str[i]+1); } std::cout << "String A Is Encrypted Now With This Format: " << "\n"; for(i=0;((i<10) && (str[i]!='\0'));i++) { std::cout << str2[i]; } std::cout << "\n"; } void strencypt(char str[]) { char star=''; char str2[10]; int i; for(i=0;((i<10) && (str[i]!='\0'));i++) { if ((int)(str[i])==32) str2[i]= star; else str2[i]=(int)(str[i]+1); } std::cout << "String A Is Encrypted Now With This Format: " << "\n"; for(i=0;((i<10) && (str[i]!='\0'));i++) { std::cout << str2[i]; } std::cout << "\n"; } __________________ Personal Portfolio TecBrain Support Forum Linux VS Windows ... Dont Even Think of it .. Distribution: Slackware if (OS==Linux) return success There are 10 kinds of people, those who can read binary numbers and those who can't.

Apr 13th, 2005, 3:16 AM	#3
TecBrain Hobbyist Programmer Join Date: Sep 2004 Location: Cyprus Posts: 147 Rep Power: 4	I have tried that before and I ended up with the same results, The loop never stops ..... (Toggle Plain Text) void strencypt(char str[]) { char star=''; char str2[10]; int i; for(i=0;((i<10) && (str[i]!='\0'));i++) { if((str[i])== ' ') str2[i]= star; else str2[i]=int(str[i]+1); } std::cout << "String A Is Encrypted Now With This Format: " << "\n"; for(i=0;((i<10) && (str[i]!='\0'));i++) { std::cout << str2[i]; } std::cout << "\n"; } void strencypt(char str[]) { char star=''; char str2[10]; int i; for(i=0;((i<10) && (str[i]!='\0'));i++) { if((str[i])== ' ') str2[i]= star; else str2[i]=int(str[i]+1); } std::cout << "String A Is Encrypted Now With This Format: " << "\n"; for(i=0;((i<10) && (str[i]!='\0'));i++) { std::cout << str2[i]; } std::cout << "\n"; } __________________ Personal Portfolio TecBrain Support Forum Linux VS Windows ... Dont Even Think of it .. Distribution: Slackware if (OS==Linux) return success There are 10 kinds of people, those who can read binary numbers and those who can't.

Apr 13th, 2005, 3:43 AM	#4
kirkl_uk Programmer Join Date: Apr 2005 Location: England Posts: 86 Rep Power: 4	Why do you limit str2 to 10 chars? I think I need more info to help on this one. MSN me if you like kirkl_uk@hotmail.com Last edited by kirkl_uk; Apr 13th, 2005 at 3:56 AM. Reason: Added information

Apr 13th, 2005, 4:39 AM	#5
TecBrain Hobbyist Programmer Join Date: Sep 2004 Location: Cyprus Posts: 147 Rep Power: 4	The Problem is not the size of the array!!!! its to replace the space with a ' * '!!!!! __________________ Personal Portfolio TecBrain Support Forum Linux VS Windows ... Dont Even Think of it .. Distribution: Slackware if (OS==Linux) return success There are 10 kinds of people, those who can read binary numbers and those who can't.

Apr 13th, 2005, 6:35 AM	#7
TecBrain Hobbyist Programmer Join Date: Sep 2004 Location: Cyprus Posts: 147 Rep Power: 4	Is it!!! I dont see any hard part on it, However; The problem with this code is the loop never stops when it reaches the space character. am not sure if the problem is my compiler This is the output of entering the string "hell o" (Toggle Plain Text) EXECUTING: /home/moderator/C++/project2/main/readstri ---------------------------------------------- Enter 1 : To Read In A String Enter 2 : To Find The Size Of The String Enter 3 : To Copy String A into Sting B Enter 4 : To Merger String A With Sting B Enter 5 : To Compare Two Strings Enter 6 : To Encrypt String A, Method A Enter 7 : To Encrypt String A, Method B Enter 8 : To Locate a Character Enter 9 : To Exit The Program 6 Enter A String hell o String A Is Encrypted Now With This Format: ifmm Enter 1 : To Read In A String Enter 2 : To Find The Size Of The String Enter 3 : To Copy String A into Sting B Enter 4 : To Merger String A With Sting B Enter 5 : To Compare Two Strings Enter 6 : To Encrypt String A, Method A Enter 7 : To Encrypt String A, Method B Enter 8 : To Locate a Character Enter 9 : To Exit The Program 6 Enter A String hell o String A Is Encrypted Now With This Format: ifmm Enter 1 : To Read In A String Enter 2 : To Find The Size Of The String Enter 3 : To Copy String A into Sting B Enter 4 : To Merger String A With Sting B Enter 5 : To Compare Two Strings Enter 6 : To Encrypt String A, Method A Enter 7 : To Encrypt String A, Method B Enter 8 : To Locate a Character Enter 9 : To Exit The Program 6 Enter A String hell o String A Is Encrypted Now With This Format: ifmm Enter 1 : To Read In A String Enter 2 : To Find The Size Of The String Enter 3 : To Copy String A into Sting B Enter 4 : To Merger String A With Sting B Enter 5 : To Compare Two Strings Enter 6 : To Encrypt String A, Method A Enter 7 : To Encrypt String A, Method B Enter 8 : To Locate a Character Enter 9 : To Exit The Program 6 Enter A String hell o String A Is Encrypted Now With This Format: ifmm . . . . . . EXECUTING: /home/moderator/C++/project2/main/readstri ---------------------------------------------- Enter 1 : To Read In A String Enter 2 : To Find The Size Of The String Enter 3 : To Copy String A into Sting B Enter 4 : To Merger String A With Sting B Enter 5 : To Compare Two Strings Enter 6 : To Encrypt String A, Method A Enter 7 : To Encrypt String A, Method B Enter 8 : To Locate a Character Enter 9 : To Exit The Program 6 Enter A String hell o String A Is Encrypted Now With This Format: ifmm Enter 1 : To Read In A String Enter 2 : To Find The Size Of The String Enter 3 : To Copy String A into Sting B Enter 4 : To Merger String A With Sting B Enter 5 : To Compare Two Strings Enter 6 : To Encrypt String A, Method A Enter 7 : To Encrypt String A, Method B Enter 8 : To Locate a Character Enter 9 : To Exit The Program 6 Enter A String hell o String A Is Encrypted Now With This Format: ifmm Enter 1 : To Read In A String Enter 2 : To Find The Size Of The String Enter 3 : To Copy String A into Sting B Enter 4 : To Merger String A With Sting B Enter 5 : To Compare Two Strings Enter 6 : To Encrypt String A, Method A Enter 7 : To Encrypt String A, Method B Enter 8 : To Locate a Character Enter 9 : To Exit The Program 6 Enter A String hell o String A Is Encrypted Now With This Format: ifmm Enter 1 : To Read In A String Enter 2 : To Find The Size Of The String Enter 3 : To Copy String A into Sting B Enter 4 : To Merger String A With Sting B Enter 5 : To Compare Two Strings Enter 6 : To Encrypt String A, Method A Enter 7 : To Encrypt String A, Method B Enter 8 : To Locate a Character Enter 9 : To Exit The Program 6 Enter A String hell o String A Is Encrypted Now With This Format: ifmm . . . . . . The loop wont stop running, if you complie it you will see what I mean (Toggle Plain Text) void strencypt(char str[]) { char star=''; char str2[10]; int i; for(i=0;((i<10) && (str[i]!='\0'));i++) { if((str[i])== ' ') str2[i]= star; else str2[i]=int(str[i]+1); } std::cout << "String A Is Encrypted Now With This Format: " << "\n"; for(i=0;((i<10) && (str[i]!='\0'));i++) { std::cout << str2[i]; } std::cout << "\n"; } void strencypt(char str[]) { char star=''; char str2[10]; int i; for(i=0;((i<10) && (str[i]!='\0'));i++) { if((str[i])== ' ') str2[i]= star; else str2[i]=int(str[i]+1); } std::cout << "String A Is Encrypted Now With This Format: " << "\n"; for(i=0;((i<10) && (str[i]!='\0'));i++) { std::cout << str2[i]; } std::cout << "\n"; } and this is the part is on my main body (Toggle Plain Text) case 6: { std::cout << "Enter A String" << "\n"; std::cin >> name; strencypt(name); break; case 6: { std::cout << "Enter A String" << "\n"; std::cin >> name; strencypt(name); break; __________________ Personal Portfolio TecBrain Support Forum Linux VS Windows ... Dont Even Think of it .. Distribution: Slackware if (OS==Linux) return success There are 10 kinds of people, those who can read binary numbers and those who can't.

Apr 13th, 2005, 3:02 AM	#2
kirkl_uk Programmer Join Date: Apr 2005 Location: England Posts: 86 Rep Power: 4	Hoping I understand this right, why don't you loop through the string, using strlen for the counter termination, and then using something simple like "if(str[i] == ' ') str2[i] = '*';" ? I did something similar to this in a previous project where I replaced a space with a _ for writing/reading files. kirkl_uk

Apr 13th, 2005, 4:49 AM	#6
kirkl_uk Programmer Join Date: Apr 2005 Location: England Posts: 86 Rep Power: 4	I am aware of this. However, your explanation of the problem is very poor. Note how no one is able to help.

Apr 13th, 2005, 7:05 AM	#8
kirkl_uk Programmer Join Date: Apr 2005 Location: England Posts: 86 Rep Power: 4	We are here to help, so try not to get so defensive. It's clear to you how the program should work and what happens during execution, but it wasn't clear for an outsider to the program. I don't have a compiler on this machine at the moment, so I will have a look at this when I get home later tonight. Remember, I am trying to help you.

Apr 13th, 2005, 8:04 AM	#9
TecBrain Hobbyist Programmer Join Date: Sep 2004 Location: Cyprus Posts: 147 Rep Power: 4	kirkl_UK I wasn't trying to offend you, however; if I did please accept my apology. some how I think I need to use the command gets instead of cin!!!Not sure though I'll try it. __________________ Personal Portfolio TecBrain Support Forum Linux VS Windows ... Dont Even Think of it .. Distribution: Slackware if (OS==Linux) return success There are 10 kinds of people, those who can read binary numbers and those who can't. Last edited by TecBrain; Apr 13th, 2005 at 8:09 AM.

Apr 13th, 2005, 9:14 AM	#10
spydoor Programmer Join Date: Feb 2005 Posts: 64 Rep Power: 4	wierd.. you thought of the answer but don't know why. (Toggle Plain Text) cin >> name; // reads up to whitespace, newline , or EOF cin >> name; // reads up to whitespace, newline , or EOF so you were never reading in a space to begin with. (Toggle Plain Text) cin.getline(name, length); // reads up to newline, or EOF cin.getline(name, length); // reads up to newline, or EOF Last edited by spydoor; Apr 13th, 2005 at 9:21 AM.

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Programming Forums > Application Development > C++

replace space with ' * '