new and delete

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Original post by TelevirtualSoftDev Sorry, silly mistake...thanks for pointing it out though. It should read: char* cAsciiArrayPtr = NULL; cAsciiArrayPtr = arrayCopier.copyToCharArray(argv[1]); (the 'new' is in here) delete[] cAsciiArrayPtr; cAsciiArrayPtr = NULL; thanks for the other tip to...i'll look into it.

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Original post by TelevirtualSoftDev I'm feeding the char array into an API for a Text to Speech engine which doesn't like it.

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Original post by richardurich Doesn't string::c_str return a const char*? If so, it is not equivalent to code which returns a char*.

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Original post by snk_kid For all other cases you will need to make a separate copy but TelevirtualSoftDev is reinventing wheels anways since you can/should use std::copy or at very least std::strcpy/strncpy.

  #include <cstring>  #include <vector>  // ...  std::vector<char> asciiArray(argv[1], argv[1]+std::strlen(argv[1]));  legacyCallTaskingCharPointers(&asciiArray[0]);

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Original post by Bregma Quote: Original post by snk_kid For all other cases you will need to make a separate copy but TelevirtualSoftDev is reinventing wheels anways since you can/should use std::copy or at very least std::strcpy/strncpy. Nay, for all other cases you should used vector. For gosh sake, if all you need is a vector of type char, use a vector of type char and stick with the native idioms. #include <cstring> #include <vector> // ... std::vector<char> asciiArray(argv[1], argv[1]+std::strlen(argv[1])); legacyCallTaskingCharPointers(&asciiArray[0]);

#include <string>#include <vector>#include <algorithm>// .....std::string s = argv[1];// ....std::vector<char> cs(s.length() + 1);std::copy(s.begin(), s.end(), cs.begin());cs[s.length()] = '\0';legacyAPIFunction(&cs[0]);

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Original post by snk_kid Quote: Original post by Bregma Quote: Original post by snk_kid For all other cases you will need to make a separate copy but TelevirtualSoftDev is reinventing wheels anways since you can/should use std::copy or at very least std::strcpy/strncpy. Nay, for all other cases you should used vector. For gosh sake, if all you need is a vector of type char, use a vector of type char and stick with the native idioms. Yes good point however you could make it better time complexity wise if you use std::string from the begining.

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Original post by snk_kid Yes good point however you could make it better time complexity wise if you use std::string from the begining. #include <string>#include <vector>// Allocates O(strlen(argv[1])) chars, copies strlen(argv[1]) chars from argv[1] into s.std::string s = argv[1];// Allocates s.length()+1 chars, initializaes them to 0.std::vector<char> cs(s.length() + 1);// Copies s.length() chars from s into cs.std::copy(s.begin(), s.end(), cs.begin());// Sets the EOS. Note it's already zero.cs[s.length()] = '\0';legacyAPIFunction(&cs[0]);

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Original post by snk_kid Quote: Original post by Bregma Quote: Original post by snk_kid For all other cases you will need to make a separate copy but TelevirtualSoftDev is reinventing wheels anways since you can/should use std::copy or at very least std::strcpy/strncpy. Nay, for all other cases you should used vector. For gosh sake, if all you need is a vector of type char, use a vector of type char and stick with the native idioms. #include <cstring> #include <vector> #include <algorithm> // ... std::vector<char> asciiArray(argv[1], argv[1]+std::strlen(argv[1])); legacyCallTaskingCharPointers(&asciiArray[0]); Yes good point however you could make it better time complexity wise if you use std::string from the begining. #include <string>#include <vector>// .....std::string s = argv[1];// ....std::vector<char> cs(s.length() + 1);std::copy(s.begin(), s.end(), cs.begin());cs[s.length()] = '\0';legacyAPIFunction(&cs[0]);

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Original post by Zahlman Uh, they both look O(n) to me :s

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Original post by Bregma Hmm, sounds like twice as many memory allocations and copies over just using the vector.

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Original post by Bregma I would suggest that creating a vector and then using copy is more work (for me) than just creating the vector using the range constructor. My time is money, and more work means higher cost.

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Original post by snk_kid Quote: Original post by Zahlman Uh, they both look O(n) to me :s I'm alluding to many things here, look closely at strlen if you're not going to maintain this vector of chars and you're going to do this on a regular basis you're paying an extra O(n) for strlen each time because strlen is not a constant time operation, where as my example is sacrificing a minute & insignicant amount of memory, allocated just once (instead of using argv) for time. This is what the "..." was implying. if you're just doing this on a one-off then it's not worth it.

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Original post by ordered_disorder One strlen, one allocation and one copy, doesn't seem as costly as two copies, two allocations and the overhead of a vector and string. We're talking miniscule overhead in either case, but IF we are going to make the comparisons. There's also the readability coefficient. *EDITED FOR POOR WORDING*

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Original post by snk_kid Since you're not bothering to read carefully i'm not going to bother acknowledging your comments.

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Original post by snk_kid Quote: Original post by Zahlman Uh, they both look O(n) to me :s I'm alluding to many things here, look closely at strlen if you're not going to maintain this vector of chars and you're going to do this on a regular basis you're paying an extra O(n) for strlen each time because strlen is not a constant time operation, where as my example is sacrificing a minute & insignicant amount of memory, allocated just once (instead of using argv) for time. This is what the "..." was implying. if you're just doing this on a one-off then it's not worth it.

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Original post by snk_kid Quote: Original post by Bregma I would suggest that creating a vector and then using copy is more work (for me) than just creating the vector using the range constructor. My time is money, and more work means higher cost. There is no need to be so condescending, beside the point what you said about saving time is a little exaggerated because if that was the case you wouldn't be using C++ to begin with. Are you implying that C++ is a highly prodcutive & elegant langauge? well it isn't.