Programming Forums > Application Development > C++

myName · Feb 23rd, 2006, 9:55 PM

I have a a string

char *chMsg = "02222AAA00";

i wana separate this "02222AAA00" into part by part..
Like this "0", "2222", "AAA", "00".

so i declared 4 variables

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char *chOne = new char[1]; //"0"
char *chTwo = new char[4]; //"2222"
char *chThree= new char[3]; //"AAA"
char *chFour = new char[2]; //"00"

than,
i use strncpy

(Toggle Plain Text)

strncpy(chOne, chMsg, 1);

to get the "0" from "02222AAA00".

but how to get the "2222" from "02222AAA00"???

myName · Feb 23rd, 2006, 10:32 PM

i try this

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             string strMsg = chMsg;
	string strOne = strMsg.substr(0,1);
	string strTwo = strMsg.substr(1,4);
	string strThree = strMsg.substr(5,3);
	string strFour =strMsg.substr(8,2);

	printf( "chMsg: %s \n", strMsg);

error in printf, the use of %s.
i found that %s is just use for char *.
How bout string data type?

Kaja Fumei · Feb 23rd, 2006, 10:53 PM

You can use the c_str() method in the string type to return the string as a char*:

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printf( "chMsg: %s \n", strMsg.c_str());

nnxion · Feb 24th, 2006, 2:28 AM

Since this is in the C++ forum, just use:

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std::cout << strMsg << std::endl;

grumpy · Feb 24th, 2006, 4:35 AM

Quote:

Originally Posted by myName

I have a a string

(Toggle Plain Text)

char *chMsg = "02222AAA00";

i wana separate this "02222AAA00" into part by part..
Like this "0", "2222", "AAA", "00".

so i declared 4 variables

(Toggle Plain Text)

	
char *chOne = new char[1]; //"0"
char *chTwo = new char[4]; //"2222"
char *chThree= new char[3]; //"AAA"
char *chFour = new char[2]; //"00"

The lengths of each of the new'd arrays are one too short (particularly if you want to treat chOne etc as strings without invoking undefined behaviour).

C-strings include a trailing zero byte, so you need to do this;

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char *chOne = new char[2]; //"0"
char *chTwo = new char[5]; //"2222"
char *chThree= new char[4]; //"AAA"
char *chFour = new char[3]; //"00"

Quote:

Originally Posted by myName

than,
i use strncpy

(Toggle Plain Text)

strncpy(chOne, chMsg, 1);

to get the "0" from "02222AAA00".

This is technically OK, unless you attempt to use chOne as a string. For example;

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fprintf(stdout, "%s\n", chOne);

will generate undefined behaviour (in practice, a common result will be garbage output, but anything is actually allowed to happen).

To eliminate the undefined behaviour, do this immediately after the above strncpy() call....

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chOne[1] = '\0';

Quote:

Originally Posted by myName

but how to get the "2222" from "02222AAA00"???

I'm assuming you can assume, as in this example, that the "2222" substring starts at the second byte in the string (i.e. at index 1) and you know it is of length 4. In that case, the approach is;

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   strncpy(chTwo, chMsg + 1, 4);
   chTwo[4] = '\0';

myName · Feb 26th, 2006, 7:37 PM

Kaja Fumei, nnxion and, grumpy,

Thanks guys. Very appreciate all yours help. :-)
Grumpy, your explained very detail and clear, make me very easy to understand the concepts..

Thanks all. :-)

titaniumdecoy · Feb 26th, 2006, 9:10 PM

I'm not so good at C++, but I thought the char type was for storing single characters, not strings, such as char *chMsg = "02222AAA00". How does that work?

Jimbo · Feb 26th, 2006, 9:55 PM

a char is just a single character, while a char* (which could also be written as char []) is an array of characters, hence the "02222AAA00".

titaniumdecoy · Feb 26th, 2006, 10:56 PM

I thought a * meant that a variable was a pointer... (to a single char)?

Jimbo · Feb 26th, 2006, 11:19 PM

A pointer points to a char in memory, but an array is a sequence of consecutive chars in mem. So you could have a pointer to the first one, and then iterate through memory (by incrementing the pointer) to get the rest of the array.

Basically, arrays are pointers that only point the the first item yet still index the whole array. (array[i] == *(array+i))

You'll see the similarities a lot more when you use dynamic memory allocation

Feb 23rd, 2006, 9:55 PM	#1
myName Programmer Join Date: Oct 2005 Posts: 48 Rep Power: 0	char * I have a a string (Toggle Plain Text) char chMsg = "02222AAA00"; char chMsg = "02222AAA00"; i wana separate this "02222AAA00" into part by part.. Like this "0", "2222", "AAA", "00". so i declared 4 variables (Toggle Plain Text) char chOne = new char[1]; //"0" char chTwo = new char[4]; //"2222" char chThree= new char[3]; //"AAA" char chFour = new char[2]; //"00" char chOne = new char[1]; //"0" char chTwo = new char[4]; //"2222" char chThree= new char[3]; //"AAA" char chFour = new char[2]; //"00" than, i use strncpy (Toggle Plain Text) strncpy(chOne, chMsg, 1); strncpy(chOne, chMsg, 1); to get the "0" from "02222AAA00". but how to get the "2222" from "02222AAA00"???

Feb 23rd, 2006, 10:32 PM	#2
myName Programmer Join Date: Oct 2005 Posts: 48 Rep Power: 0	i try this (Toggle Plain Text) string strMsg = chMsg; string strOne = strMsg.substr(0,1); string strTwo = strMsg.substr(1,4); string strThree = strMsg.substr(5,3); string strFour =strMsg.substr(8,2); printf( "chMsg: %s \n", strMsg); string strMsg = chMsg; string strOne = strMsg.substr(0,1); string strTwo = strMsg.substr(1,4); string strThree = strMsg.substr(5,3); string strFour =strMsg.substr(8,2); printf( "chMsg: %s \n", strMsg); error in printf, the use of %s. i found that %s is just use for char *. How bout string data type?

Feb 23rd, 2006, 10:53 PM	#3
Kaja Fumei Hobbyist Programmer Join Date: Oct 2005 Posts: 134 Rep Power: 3	You can use the c_str() method in the string type to return the string as a char*: (Toggle Plain Text) printf( "chMsg: %s \n", strMsg.c_str()); printf( "chMsg: %s \n", strMsg.c_str());

Feb 24th, 2006, 2:28 AM	#4
nnxion Programming Guru Join Date: Jun 2005 Location: elemental plane Posts: 1,429 Rep Power: 5	Since this is in the C++ forum, just use: (Toggle Plain Text) std::cout << strMsg << std::endl; std::cout << strMsg << std::endl; __________________ "Employ your time in improving yourself by other men's writings, so that you shall gain easily what others have labored hard for." -- Socrates

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Feb 26th, 2006, 7:37 PM	#6
myName Programmer Join Date: Oct 2005 Posts: 48 Rep Power: 0	Kaja Fumei, nnxion and, grumpy, Thanks guys. Very appreciate all yours help. :-) Grumpy, your explained very detail and clear, make me very easy to understand the concepts.. Thanks all. :-)

Feb 26th, 2006, 9:10 PM	#7
titaniumdecoy Expert Programmer Join Date: Nov 2005 Posts: 837 Rep Power: 3	I'm not so good at C++, but I thought the char type was for storing single characters, not strings, such as char *chMsg = "02222AAA00". How does that work?

Feb 26th, 2006, 9:55 PM	#8
Jimbo Battle Programmer Join Date: Feb 2006 Location: Bellevue, WA, USA Posts: 746 Rep Power: 3	a char is just a single character, while a char* (which could also be written as char []) is an array of characters, hence the "02222AAA00".

Feb 26th, 2006, 10:56 PM	#9
titaniumdecoy Expert Programmer Join Date: Nov 2005 Posts: 837 Rep Power: 3	I thought a * meant that a variable was a pointer... (to a single char)?

Feb 26th, 2006, 11:19 PM	#10
Jimbo Battle Programmer Join Date: Feb 2006 Location: Bellevue, WA, USA Posts: 746 Rep Power: 3	A pointer points to a char in memory, but an array is a sequence of consecutive chars in mem. So you could have a pointer to the first one, and then iterate through memory (by incrementing the pointer) to get the rest of the array. Basically, arrays are pointers that only point the the first item yet still index the whole array. (array[i] == *(array+i)) You'll see the similarities a lot more when you use dynamic memory allocation

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