Convert char array to char*

Probably another stupid question :) I need to convert an array of char to a single char*. But not all elements - I want to ignore the first and last element. So, if char digits[6] contains the following:

digits[0] = "A";
digits[1] = "B";
digits[2] = "C";
digits[3] = "D";
digits[4] = "E";
digits[5] = "F";

then I want char * alldigits to contain "BCDE". I have spent ages looking for a solution to this, but can't find anything that works!

TIA

You'll have to copy the characters into a new string, as the last element in a 'C' string must be '\0'. Doing this to your original array would require you to overwrite the 'F' with '\0'.

Ok, I can make it a string:

Code:

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string alldigits("");
alldigits += digits[1];
alldigits += digits[2];
alldigits += digits[3];
alldigits += digits[4];

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But how do I then make that a char*?

Quote:

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Originally Posted by motorollin

Ok, I can make it a string:

Code:

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string alldigits("");
alldigits += digits[1];
alldigits += digits[2];
alldigits += digits[3];
alldigits += digits[4];

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But how do I then make that a char*?

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First, the terminology is incorrect.

A char* is a pointer. A std::string is a class that represents a character string. They are not the same thing. A char* may point to the start of a character buffer (or just a single char), depending on how it's used.

Second, The std::string.c_str() returns a const char * that represents the data of the std::string.

Third, if you plan on changing what the char* points to, you cannot do it with std::string::c_str(), as the pointer returned is const. Either use a regular char array, use std::vector<char>, or use new[]/delete[] for routines that explicitly want a char * (not const char *).

Regards,

Paul McKenzie

Something along the lines of:

Code:

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#include <iostream>
#include <vector>

size_t getSize(std::string str){return str.size();}

int main()
{
  char* digits = "ABCDEF";
  size_t size = getSize(digits);

  std::vector<char> vec;
  if (size > 2)
    vec.assign(digits+1, digits+size-1);
  vec.push_back('\0');

  char* result = &vec[0];
 
  std::cout << result << std::endl;
  system("PAUSE");
}

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If you don't mind modifying the original data, the simplest way is:

Code:

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char digits[6];
char *ptr;

digits[0] = "A";
digits[1] = "B";
digits[2] = "C";
digits[3] = "D";
digits[4] = "E";
digits[5] = "F";

ptr = &digits[1];
digits[5] = 0;

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Also note that while a char array and a char* are convertible in 1D, the same is *not* true of 2D arrays----you can't convert a char[10][10] into a char**, for instance. This is because of the way the sizeof() operator handles each one....

Quote:

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Originally Posted by motorollin

Probably another stupid question :) I need to convert an array of char to a single char*. But not all elements - I want to ignore the first and last element. So, if char digits[6] contains the following:

digits[0] = "A";
digits[1] = "B";
digits[2] = "C";
digits[3] = "D";
digits[4] = "E";
digits[5] = "F";

then I want char * alldigits to contain "BCDE". I have spent ages looking for a solution to this, but can't find anything that works!

TIA

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chars are represented by single quotes. The code you posted shouldn't compile.

One way to get what you want would be

char digits2[5];
strncpy(digits2, digits + 1, 4);
digits2[4] = 0;

^You would of course need to add
digits2[4] = 0;
for that to work.

Thanks for the help guys. I really don't know what to do with this, and to be honest that stems from not understanding the various data types in C++. I'll go and do some reading about that and then come and have another look. Hopefully it will make more sense then!

Cheers

Quote:

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Originally Posted by Lindley

^You would of course need to add
digits2[4] = 0;
for that to work.

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Corrected. Thanks.