I am trying to practice C++ and while doing so I ran into a problem in my code. I dynamically create a character array and then for each array index, I want to fill that element with an integer. I tried casting the integer to a character but that didn't seem to work. After printing out the array element, nothing comes out. I would appreciate any help, I'm pretty new to this, thanks .
char *createBoard()
{
char *theGameBoard = new char[8];
for (int i = 0; i < 8; i++)
theGameBoard[i] = (char)i; //doesn't work
return theGameBoard;
}
Here is how I ended up doing it.
char *createBoard()
{
char *theGameBoard = new char[8];
theGameBoard[0] = '0';
theGameBoard[1] = '1';
theGameBoard[2] = '2';
theGameBoard[3] = '3';
theGameBoard[4] = '4';
theGameBoard[5] = '5';
theGameBoard[6] = '6';
theGameBoard[7] = '7';
theGameBoard[8] = '8';
return theGameBoard;
}
Basically, your two sections of code are not quite equivalent.
When you set theGameBoard[0] = '0' you are essentially setting it to the value 48 (the ASCII code for the character '0'). So setting theGameBoard[0] = (char)i is not quite the same thing if i = 0. You need to add the offset of '0' in the ASCII table (which is 48) so that theGameBoard[0] is actually 0 + the offset of character '0'.
Here's how you do it:
char *createBoard()
{
char *theGameBoard = new char[8];
for (int i = 0; i < 8; i++)
theGameBoard[i] = '0' + (char)i; //you want to set each array cell
// to an ASCII numnber (so use '0' as an offset)
return theGameBoard;
}
Also, like #Daniel said: make sure that you free up the memory that you are allocating in this function after you are done with using the returned variable. Like so:
int main()
{
char* gameBoard = createBoard();
// you can now use the gameBoard variable here
// ...
// when you are done with it
// make sure to call delete on it
delete[] gameBoard;
// exit the program here..
return 0;
}
Your second function has an off-by-one bug. You allocate an array of length 8, but you copy 9 values into it. (0, 1, 2, 3, 4, 5, 6, 7, 8).
If I was doing this I would use stringstream. It might be heavy weight for this, but it is the C++ way of doing things.
for (int i = 0; i < 8; ++i) {
stringstream sstream;
sstream << i;
sstream >> theGameBoard[i];
}
When you are done using the game board array you need to delete it with this command:
delete[] theGameBoard;
In your character array you must store ASCII values of digits.
For example:
ASCII value of '0' is 48 ( not 0 )
ASCII value of '1' is 49 ( not 1 )
...
In C++ ( and almost every other language ) you can get ASCII value of character putting it in single quote ( '0' == 48 , '1' == 49 , '2' == 50 , ... )
Your array must have values
theGameBoard[0] = '0'
theGameBoard[1] = '1' or theGameBoard = '0' + 1
...
Code that fills your array:
for(int k=0;k<8;++k)
theGameBoard[k] = '0' + k;
Related
I want to select the first 8 characters of a string using C++. Right now I create a temporary string which is 8 characters long, and fill it with the first 8 characters of another string.
However, if the other string is not 8 characters long, I am left with unwanted whitespace.
string message = " ";
const char * word = holder.c_str();
for(int i = 0; i<message.length(); i++)
message[i] = word[i];
If word is "123456789abc", this code works correctly and message contains "12345678".
However, if word is shorter, something like "1234", message ends up being "1234 "
How can I select either the first eight characters of a string, or the entire string if it is shorter than 8 characters?
Just use std::string::substr:
std::string str = "123456789abc";
std::string first_eight = str.substr(0, 8);
Just call resize on the string.
If I have understood correctly you then just write
std::string message = holder.substr( 0, 8 );
Jf you need to grab characters from a character array then you can write for example
const char *s = "Some string";
std::string message( s, std::min<size_t>( 8, std::strlen( s ) );
Or you could use this:
#include <climits>
cin.ignore(numeric_limits<streamsize>::max(), '\n');
If the max is 8 it'll stop there. But you would have to set
const char * word = holder.c_str();
to 8. I believe that you could do that by writing
const int SIZE = 9;
char * word = holder.c_str();
Let me know if this works.
If they hit space at any point it would only read up to the space.
char* messageBefore = "12345678asdfg"
int length = strlen(messageBefore);
char* messageAfter = new char[length];
for(int index = 0; index < length; index++)
{
char beforeLetter = messageBefore[index];
// 48 is the char code for 0 and
if(beforeLetter >= 48 && beforeLetter <= 57)
{
messageAfter[index] = beforeLetter;
}
else
{
messageAfter[index] = ' ';
}
}
This will create a character array of the proper size and transfer over every numeric character (0-9) and replace non-numerics with spaces. This sounds like what you're looking for.
Given what other people have interpreted based on your question, you can easily modify the above approach to give you a resulting string that only contains the numeric portion.
Something like:
int length = strlen(messageBefore);
int numericLength = 0;
while(numericLength < length &&
messageBefore[numericLength] >= 48 &&
messageBefore[numericLength] <= 57)
{
numericLength++;
}
Then use numericLength in the previous logic in place of length and you'll get the first bunch of numeric characters.
Hope this helps!
How could I change an array of character (string) to an integer number without using any ready function (ex atoi();) for example :-
char a[5]='4534';
I want the number 4534 , How can I get it ?
Without using any existing libraries, you have to:
Convert character to numeric digit.
Combine digits to form a number.
Converting character to digit:
digit = character - '0';
Forming a number:
number = 0;
Loop:
number = number * 10 + digit;
Your function will have to check for '+' and '-' and other non-digits characters.
First of all this statement
char a[5]='4534';
will not compile. I think you mean
char a[5]="4534";
^^ ^^
To convert this string to a number is enough simple.
For example
int number = 0;
for ( const char *p = a; *p >= '0' && *p <= '9'; ++p )
{
number = 10 * number + *p - '0';
}
Or you could skip leading white spaces.
For example
int number = 0;
const char *p = s;
while ( std::isspace( ( unsigned char )*p ) ) ++p;
for ( ; *p >= '0' && *p <= '9'; ++p )
{
number = 10 * number + *p - '0';
}
If the string may contain sign '+' or '-' then you can check at first whether the first character is a sign.
You can go like this
It's working
#include <stdio.h>
#include <string.h>
#include <math.h>
int main()
{
char a[5] = "4534";
int converted = 0;
int arraysize = strlen(a);
int i = 0;
for (i = 0; i < arraysize ; i++)
{
converted = converted *10 + a[i] - '0';
}
printf("converted= %d", converted);
return 0;
}
I prefer to use the std::atoi() function to convert string into a number data type. In your example you have an non-zero terminated array "\0", so the function will not work with this code.
Consider to use zero terminated "strings", like:
char *a = "4534";
int mynumber = std::atoi( a ); // mynumber = 4534
How could I change an array of character (string) to an integer number without using any ready function (ex atoi();) for example :-
char a[5]='4534';
I want the number 4534 , How can I get it ?
Without using any existing libraries, you have to:
Convert character to numeric digit.
Combine digits to form a number.
Converting character to digit:
digit = character - '0';
Forming a number:
number = 0;
Loop:
number = number * 10 + digit;
Your function will have to check for '+' and '-' and other non-digits characters.
First of all this statement
char a[5]='4534';
will not compile. I think you mean
char a[5]="4534";
^^ ^^
To convert this string to a number is enough simple.
For example
int number = 0;
for ( const char *p = a; *p >= '0' && *p <= '9'; ++p )
{
number = 10 * number + *p - '0';
}
Or you could skip leading white spaces.
For example
int number = 0;
const char *p = s;
while ( std::isspace( ( unsigned char )*p ) ) ++p;
for ( ; *p >= '0' && *p <= '9'; ++p )
{
number = 10 * number + *p - '0';
}
If the string may contain sign '+' or '-' then you can check at first whether the first character is a sign.
You can go like this
It's working
#include <stdio.h>
#include <string.h>
#include <math.h>
int main()
{
char a[5] = "4534";
int converted = 0;
int arraysize = strlen(a);
int i = 0;
for (i = 0; i < arraysize ; i++)
{
converted = converted *10 + a[i] - '0';
}
printf("converted= %d", converted);
return 0;
}
I prefer to use the std::atoi() function to convert string into a number data type. In your example you have an non-zero terminated array "\0", so the function will not work with this code.
Consider to use zero terminated "strings", like:
char *a = "4534";
int mynumber = std::atoi( a ); // mynumber = 4534
Below is an example code that is not working the way I want.
#include <iostream>
using namespace std;
int main()
{
char testArray[] = "1 test";
int numReplace = 2;
testArray[0] = (int)numReplace;
cout<< testArray<<endl; //output is "? test" I wanted it 2, not a '?' there
//I was trying different things and hoping (int) helped
testArray[0] = '2';
cout<<testArray<<endl;//"2 test" which is what I want, but it was hardcoded in
//Is there a way to do it based on a variable?
return 0;
}
In a string with characters and integers, how do you go about replacing numbers? And when implementing this, is it different between doing it in C and C++?
If numReplace will be in range [0,9] you can do :-
testArray[0] = numReplace + '0';
If numReplace is outside [0,9] you need to
a) convert numReplace into string equivalent
b) code a function to replace a part of string by another evaluated in (a)
Ref: Best way to replace a part of string by another in c and other relevant post on SO
Also, since this is C++ code, you might consider using std::string, here replacement, number to string conversion, etc are much simpler.
You should look over the ASCII table over here: http://www.asciitable.com/
It's very comfortable - always look on the Decimal column for the ASCII value you're using.
In the line: TestArray[0] = (int)numreplace; You've actually put in the first spot the character with the decimal ASCII value of 2. numReplace + '0' could do the trick :)
About the C/C++ question, it is the same in both and about the characters and integers...
You should look for your number start and ending.
You should make a loop that'll look like this:
int temp = 0, numberLen, i, j, isOk = 1, isOk2 = 1, from, to, num;
char str[] = "asd 12983 asd";//will be added 1 to.
char *nstr;
for(i = 0 ; i < strlen(str) && isOk ; i++)
{
if(str[i] >= '0' && str[i] <= '9')
{
from = i;
for(j = i ; j < strlen(str) && isOk2)
{
if(str[j] < '0' || str[j] > '9')//not a number;
{
to=j-1;
isOk2 = 0;
}
}
isOk = 0; //for the loop to stop.
}
}
numberLen = to-from+1;
nstr = malloc(sizeof(char)*numberLen);//creating a string with the length of the number.
for(i = from ; i <= to ; i++)
{
nstr[i-from] = str[i];
}
/*nstr now contains the number*/
num = atoi(numstr);
num++; //adding - we wanted to have the number+1 in string.
itoa(num, nstr, 10);//putting num into nstr
for(i = from ; i <= to ; i++)
{
str[i] = nstr[i-from];
}
/*Now the string will contain "asd 12984 asd"*/
By the way, the most efficient way would probably be just looking for the last digit and add 1 to it's value (ASCII again) as the numbers in ASCII are following each other - '0'=48, '1'=49 and so on. But I just showed you how to treat them as numbers and work with them as integers and so. Hope it helped :)
So I have a program that makes char* stuff lowercase. It does it by iterating through and manipulating the ascii. Now I know there's probably some library for this in c++, but that's not the point - I'm a student trying to get a grasp on char*s and stuff :).
Here's my code:
#include <iostream>
using namespace std;
char* tolower(char* src);
int main (int argc, char * const argv[])
{
char* hello = "Hello, World!\n";
cout << tolower(hello);
return 0;
}
char* tolower(char* src)
{
int ascii;
for (int n = 0; n <= strlen(src); n++)
{
ascii = int(src[n]);
if (ascii >= 65 && ascii <= 90)
{
src[n] = char(ascii+32);
}
}
return src;
}
( this is not for an assignment ;) )
It builds fine, but when I run it it I get a "The Debugger has exited due to signal 10" and Xcode points me to the line: "src[n] = char(ascii+32);"
Thanks!
Mark
Yowsers!
Your "Hello World!" string is what is called a string literal, this means its memory is part of the program and cannot be written to.
You are performing what is called an "in-place" transform, e.g. instead of writing out the lowercase version to a new buffer you are writing to the original destination. Because the destination is a literal and cannot be written to you are getting a crash.
Try this;
char hello[32];
strcpy(hello, "Hello, World!\n");
Also in your for loop, you should use <, not <=. strlen returns the length of a string minus its null terminator, and array indices are zero-based.
As Andrew noted "Hello World\n" in code is a read-only literal. You can either use strcpy to make a modifiable copy, or else try this:
char hello[] = "Hello, World!\n";
This automatically allocates an array on the stack big enough to hold a copy of the literal string and a trailing '\0', and copies the literal into the array.
Also, you can just leave ascii as a char, and use character literals instead of having to know what the numeric value of 'A' is:
char ascii;
for (int n = 0; n < strlen(src); n++)
{
ascii = src[n];
if (ascii >= 'A' && ascii <= 'Z')
{
src[n] = ascii - 'A' + 'a';
}
}
While you're at it, why bother with ascii at all, just use src[n]:
for (int n = 0; n < strlen(src); n++)
{
if (src[n] >= 'A' && src[n] <= 'Z')
{
src[n] -= 'A' - 'a';
}
}
And then, you can take advantage of the fact that in order to determine the length of a c-string, you have to iterate though it anyway, and just combine both together:
for (char *n = src; *n != 0; n++)
if (*n >= 'A' && *n <= 'Z')
*n -= 'A' - 'a';