Thread: Convert

Originally Posted by Ancient Dragon

...but there is no guarantee that the digits have consecutive ascii values. for example, the following code is not portable.

yes there is. the ASCII chart is what it is. If your system uses ASCII, you're guaranteed that 65 = 'A', 66 = 'B', and 0x30-0x39 are 1-9. so long as your system uses ASCII, '0' will always be equal to (char)48

and atoi() is less portable - it's not even standard code.

as for the OP's question, you can just walk through your array, and cast each one. for example, the most portable way, is in fact, to rely on the ASCII table:

Code:

#include<iostream>	//for console I/O
#include<cstring>	//for strlen

int main()	//main program entrypoint
{
	char letters[]="12345";			//a character array
	const int ARRLEN=strlen(letters);	//the length of the array
	int*numbers=new int[ARRLEN];		//an int array the same length
	
	for(int i=0;i<ARRLEN;i++)		//loop through the array
	{
		numbers[i]=static_cast<int>(letters[i]-'0');	//make a cast
		std::cout<<numbers[i]<<", ";	//output the results
	}

	std::cout<<std::endl;			//newline and buffer flush
	return 0;				//return to the OS
}

In that example, I used a variable size on the integer array because that frees you up to change the size of the character array and not worry about breaking your code.

Originally Posted by major_small

yes there is. the ASCII chart is what it is. If your system uses ASCII,.

you just proved my point -- not every system uses ASCII formats, which means numeric digits may (or may not) be consecutive.

Originally Posted by major_small

and atoi() is less portable - it's not even standard code

Yes it is -- see this

yeah, but it's a whole lot more portable than using atoi(). whether you're using ascii or not is system dependant, whereas whether you can use atoi() is compiler-dependant. and you can always test the system to see what character coding it's using, and adjust during runtime to suit the system. for example, you could have a simple loop that casts the letters '0'..'9' to integers. if the integers are consecutive, then you can use that method. If not, you can have your program create a lookup table, or find some other method.

edit: in writing that, I realized what I wrote before wasn't the most portable way... I think this is:

Code:

#include<iostream>	//for console I/O
#include<cstring>	//for strlen

int findint(const char letter);

int main()	//main program entrypoint
{
	char letters[]="12345";			//a character array
	const int ARRLEN=strlen(letters);	//the length of the array
	int*numbers=new int[ARRLEN];		//an int array the same length
	
	for(int i=0;i<ARRLEN;i++)		//loop through the array
	{
		numbers[i]=findint(letters[i]);	//make a cast
		std::cout<<numbers[i]<<", ";	//output the results
	}

	std::cout<<std::endl;			//newline and buffer flush
	return 0;				//return to the OS
}

int findint(const char letter)
{
	if(letter=='0')
		return 0;
	else if(letter=='1')
		return 1;
	else if(letter=='2')
		return 2;
	else if(letter=='3')
		return 3;
	else if(letter=='4')
		return 4;
	else if(letter=='5')
		return 5;
	else if(letter=='6')
		return 6;
	else if(letter=='7')
		return 7;
	else if(letter=='8')
		return 8;
	else if(letter=='9')
		return 9;

	return -1;
}

Originally Posted by Ancient Dragon

Yes it is -- see this

I knew that

Originally Posted by Dante Shamest

I think it is. Perhaps you meant itoa()?

Yeah, I was thinking about itoa(), and didn't bother to look up atoi()... I got thrown off yesterday because I'd always used atoi as standard code, and somebody pointed out that itoa (they said "itoa, etc") wasn't standard.

Originally Posted by Ancient Dragon

you just proved my point -- not every system uses ASCII formats, which means numeric digits may (or may not) be consecutive.

Actually, the C standard states that the character set's digits from 0 to 9 shall be consecutive.

Edit: At least this is the case in a C99 draft. Fortunately, all character sets are this way (at least this is the case for a practical value of 'all'.) :-)

Edit 2: Excuse me, I forgot what board I'm on.