Convert

I am trying to convert a char array into an int array, does anyone know how to start setting up the loop?

Jae

depends on the char array. post an example of the array, such as

Code:

char array[] = "12345";

If you want to convert each digit in that array to an integer, then the most portable way will be to put it into another null-terminated string array and use atoi() to convert it.

Code:

char array[] = "12345"; char buf[2] = {0} buf[0] = array[0]; int n = atoi(buf);

simply subtracting '0' from the digit might work on many operating sytems, but there is no guarantee that the digits have consecutive ascii values. for example, the following code is not portable.

Code:

char array[] = "12345"; int n = array[0] - '0';

Thanks

Thanks Ancient Dragon, that solved my problem.

Quote:

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.

Quote:

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.

Quote:

Originally Posted by major_small

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

Yes it is -- see this

Quote:

Originally Posted by major_small

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

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

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; }

Quote:

Originally Posted by Ancient Dragon

Yes it is -- see this

I knew that :(

Quote:

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.

Quote:

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.

>> numeric digits may (or may not) be consecutive

The C++ standard guarantees that numeric digits are consecutive. In section 2.2 paragraph 3:

Quote:

In both the source and execution basic character sets, the value of each character after 0 in the above list of decimal digits shall be one greater than the value of the previous.

which refers to this table in paragraph 1:

Code:

a b c d e f g h i j k l m n o p q r s t u v w x y z A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 0 1 2 3 4 5 6 7 8 9 _ { } [ ] # ( ) < > % : ; . ? * + - / ˆ & | ˜ ! = , \ " ’

So for converting a single character digit to its integer version, subtracting '0' is guaranteed to work in any valid character set.