C++ - How to append a char to char*?

Question

I've tried so may ways on the Internet to append a character to a char* but none of them seems to work. Here is one of my incomplete solution:

char* appendCharToCharArray(char * array, char a)
{
    char* ret = "";
    if (array!="") 
    {
        char * ret = new char[strlen(array) + 1 + 1]; // + 1 char + 1 for null;
        strcpy(ret,array);
    }
    else
    {
        ret = new char[2];
        strcpy(ret,array);
    }

    ret[strlen(array)] = a;  // (1)
    ret[strlen(array)+1] = '\0';
    return ret;
}

This only works when the passed array is "" (blank inside). Otherwise it doesn't help (and got an error at (1)). Could you guys please help me with this ? Thanks so much in advanced !

You cannot append a `char` to a `char*`; they're two very different types. You can append a `char` to the **array** that a `char*` points to. — Pete Becker, Nov 10 '13 at 17:13

masoud · Accepted Answer · 2017-07-22T10:45:56.093

Remove those char * ret declarations inside if blocks which hide outer ret. Therefor you have memory leak and on the other hand un-allocated memory for ret.

To compare a c-style string you should use strcmp(array,"") not array!="". Your final code should looks like below:

char* appendCharToCharArray(char* array, char a)
{
    size_t len = strlen(array);

    char* ret = new char[len+2];

    strcpy(ret, array);    
    ret[len] = a;
    ret[len+1] = '\0';

    return ret;
}

Note that, you must handle the allocated memory of returned ret somewhere by delete[] it.

Why you don't use std::string? it has .append method to append a character at the end of a string:

std::string str;

str.append('x');
// or
str += x;

Be careful with the leak. You need to `delete[]` the new'd variable if you no longer need it. Or better use `std::string`. — impulse, Jul 22 '17 at 08:23

score 3 · Answer 2 · edited Apr 20 '15 at 20:25

The function name does not reflect the semantic of the function. In fact you do not append a character. You create a new character array that contains the original array plus the given character. So if you indeed need a function that appends a character to a character array I would write it the following way

bool AppendCharToCharArray( char *array, size_t n, char c )
{
    size_t sz = std::strlen( array );

    if ( sz + 1 < n ) 
    {
        array[sz] = c;
        array[sz + 1] = '\0';
    }       

    return ( sz + 1 < n );
}

If you need a function that will contain a copy of the original array plus the given character then it could look the following way

char * CharArrayPlusChar( const char *array, char c )
{
    size_t sz = std::strlen( array );
    char *s = new char[sz + 2];

    std::strcpy( s, array );
    s[sz] = c;
    s[sz + 1] = '\0';

    return ( s );
}

score 1 · Answer 3 · answered Nov 10 '13 at 16:19

The specific problem is that you're declaring a new variable instead of assigning to an existing one:

char * ret = new char[strlen(array) + 1 + 1];
^^^^^^ Remove this

and trying to compare string values by comparing pointers:

if (array!="")     // Wrong - compares pointer with address of string literal
if (array[0] == 0) // Better - checks for empty string

although there's no need to make that comparison at all; the first branch will do the right thing whether or not the string is empty.

The more general problem is that you're messing around with nasty, error-prone C-style string manipulation in C++. Use std::string and it will manage all the memory allocation for you:

std::string appendCharToString(std::string const & s, char a) {
    return s + a;
}

score 0 · Answer 4 · answered Jan 12 '18 at 06:52

0

char ch = 't';
char chArray[2];
sprintf(chArray, "%c", ch);
char chOutput[10]="tes";
strcat(chOutput, chArray);
cout<<chOutput;

OUTPUT:

test

answered Jan 12 '18 at 06:52

Safa Seed

35
9

C++ - How to append a char to char*?

4 Answers4