How to push char array?

Question

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How do I use arrays in C++?

I have no idea how to push a char array.
For example if i have "HI.MY.NAME.IS" in a char array,
I would like to put one char in the middle and push the char's right to it.
So it would be something like "HI.MY.SLAME.IS" or something like that.

Any possible solutions?

Show some code, it's hard to understand what you're saying. But it sounds like you're not using C++, and you want the functions in the `` header. — Mooing Duck, Feb 14 '12 at 22:14

score 10 · Accepted Answer · edited Feb 24 '12 at 20:06

10

Use string::insert.

std::string s("test");
s.insert(s.begin()+2, 'c');

edited Feb 24 '12 at 20:06

Taryn

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answered Feb 14 '12 at 22:13

rasmus

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better to answer with a better more proper idiomatic solution than to answer a broken question correctly. This is the better solution in C++! – Feb 14 '12 at 22:21
Of course this assumes that he has access to the string header. – crush Feb 14 '12 at 22:28
i was trying to do it with char arrays, not string. – Haxify Feb 14 '12 at 22:37
but i guess this would work too. thanks! – Haxify Feb 14 '12 at 22:38

score 4 · Answer 2 · answered Feb 14 '12 at 22:13

There is no "automatic" push which relocates elements in an array. At the lowest levels, there is only read from an array element and write to an array element.

That means you need to copy each element after the insert "one index" down the array. Then you need to set the "inserted" value at its inserted index.

Library routines will do this for you without you noticing; however, you should be aware of the underlying mechanism. Otherwise, you might fall under the impression that inserting into arrays at arbitrary indexes is cheap (when it usually isn't).

score 1 · Answer 3 · answered Feb 14 '12 at 22:16

//This allocates new memory that you receive responsibility for freeing after use.
char *push(char *charArray, char *charsToPush, int pushPos) {
    char *thePushed = new char[strlen(charArray) +  strlen(charsToPush) + 1];
    memcpy(thePushed, charArray, pushPos);
    memcpy(thePushed + pushPos, charsToPush, strlen(charsToPush));
    memcpy(thePushed + pushPos + strlen(charsToPush), charArray + pushPos, strlen(charArray) - pushPos);
    thePushed[strlen(charArray) +  strlen(charsToPush)] = '\0';
    return thePushed;
}

score 1 · Answer 4 · edited May 23 '17 at 10:28

To do that you'd have to:

Create a new array that is large enough to hold the original and the new items. Arrays cannot be resized.
Copy all items from the old to the new array, leaving the places for the new characters open.
Insert the new characters.

This is quite complex. But C++ offers a simpler way - std::string:

#include <string>
#include <iostream>

int main() {
  std::string text = "HI.MY.NAME.IS";
  std::string new_text = text.substr(0, 6) + "SL" + text.substr(7, 6);
  std::cout << new_text << std::endl;
}

This program prints HI.MY.SLAME.IS. Or, even better, use insert, as @rasmus suggests.

James Michael Hare · Answer 5 · 2012-02-14T22:30:52.907

If you are limited to c-strings or require it to be replaced in-buffer, then you can do something like the below, but your buffer must be large enough to hold the modified string since it's pushing all characters down. That said, you'd be much better off using the std::string as the others suggest.

// ASSUMES buffer is large enough to store one more char
void insertAt(char* buffer, char insertMe, size_t at)
{
    size_t len = strlen(buffer);

    if (at <= len)
    {
        memcpy(buffer + at + 1, buffer + at, len - at + 1);
        buffer[at] = insertMe;
    }
}

score -1 · Answer 6 · answered Feb 14 '12 at 22:23

-1

char* string = (char*) malloc(14);
string = "HI.MY.NAME.IS";
realloc(string, 15);
for (int i = 14; i > 5; --i) {
    string[i+1] = string[i];
}

string[5] = 'S';
string[6] = 'L';

Here you go...lol

answered Feb 14 '12 at 22:23

crush

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-1 `new` and `realloc` do not mix, nor does `realloc` and data with static storage duration. – ildjarn Feb 14 '12 at 22:24
right, fixed. "nor does realloc and data with static storage duration" huh? – crush Feb 14 '12 at 22:26
1

After `string = "HI.MY.NAME.IS";`, `string` is pointing at data with static storage duration, and calling `realloc` on data with static storage duration is illegal. Maybe you meant `strcpy(string, "HI.MY.NAME.IS");` instead? – ildjarn Feb 14 '12 at 22:27
I assumed that he was receiving the "HI.MY.NAME.IS" from some input mechanism, and only put it there for example. – crush Feb 14 '12 at 22:32

How to push char array?

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