Insertion of character in an initialized array [closed] - c++

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Let the array I have declared be:
char input[] = {'a','l','d'};
Now, I have to insert another character inside this array. Can I insert my character inside this array, or do I have to change the size of the input array and then insert?
Do char input[] = {'a','l','d'} and char input[3]={'a','l','d'} both have the same meaning, or in the first case is the size of array not fixed?
What do I have to do if I want to replace all characters in my array with '3', '.', '1', '4'?

Answer 1

char input[] = {'a','l','d'} and char input[3]={'a','l','d'} do both have same meaning
Yes, they have the same meaning.
now I have to insert an another character inside this array
I have to change the size of the input array
It is not possible to change the size of an array or to insert elements into it. The size of an array remains constant through the lifetime of the array.
can u please tell me what to do to insert new character in the same array
You don't, because you can't.
In C++, I would recommend using the std::string class to represent a character string that can change size. A minimal example:
std::string input = "ald";
input.push_back('i');

Answer 2

If you have
char input[] = {'a','l','d'};
then the array is of size exactly 3, and there's no possibility to store additional characters in it.
You can overwrite:
input[1] = 'n';
but that replaces a character, not insert.
If you declare the array with some extra size, then you have some wiggle room.
char input[10] = {'a','l','d', '\0'};
Now we have room for 10 characters, and null termination to make it a proper string, so we can easily print it:
printf("input = %s\n", input);
This prints "input = ald".
Overwrite a character:
input[1] = 'n';
printf("input = %s\n", input);
This prints "input = and".
Add a character at the end:
input[3] = 'y';
input[4] = '\0'; /* since we just overwrote the old one */
printf("input = %s\n", input);
This prints "input = andy".
Slide some characters to the right, to make room for new ones
for(int i = 4; i >= 2; i--) /* move two characters, and \0, to the right */
input[i+1] = input[i];
input[2] = 'o';
printf("input = %s\n", input);
This prints "input = anody".
Another way to move characters is with memmove:
memmove(&input[4], &input[2], 4); /* move three characters, and \0, two to the right */
input[2] = 'y';
input[3] = 'b';
printf("input = %s\n", input);
This prints "input = anybody".
But this is all very fussy and error-prone. Examples like I've shown can teach you how characters, arrays, and strings work in C, but they're hard to get right, and I wouldn't necessarily recommend that you write code like this to manipulate your own strings.
If you want to replace all characters, it's easy enough, as long as you're sure there's room. Just use strcpy:
strcpy(input, "3.14");
printf("input = %s\n", input);
This prints "input = 3.14".
You have to be careful when calling strcpy, though. The destination array has to be (a) writable and (b) big enough. If you wrote
char input2[3];
strcpy(input2, "3.14"); /* WRONG */
that would be wrong, because input2 is not big enough. And if you wrote
char *input3 = "hello";
strcpy(input3, "3.14"); /* WRONG */
that would be wrong, because input3 is now a pointer, that points to an unnamed array containing the string "hello", and while it's big enough, it's probably an unwritable constant string.