C++: invalid conversion from char to char*?

Question

I receive this error:

array.cpp: In function 'void strcat2(char*, const char*)': array.cpp:74: error: invalid conversion from 'char' to 'char*' array.cpp:74: error: initializing argument 1 of 'void strcpy2(char*, const char*)' array.cpp:74: error: invalid conversion from 'char' to 'const char*' array.cpp:74: error: initializing argument 2 of 'void strcpy2(char*, const char*)'

When trying to run this code:

//Concatenates two arrays of characters using pointers
void strcat2(char* t, const char* s)
{
        unsigned int i;

        for (i = 0; *t; i++);
        strcpy2(*(t + i), *s);
}

Here is the strcpy2 function it calls:

//Copies the information from one array of characters to another using pointers
void strcpy2(char* t, const char* s)
{
        for ( ; *t++ = *s++;);
}

It says invalid conversion from char to char*, but where am I trying to convert from char to char*? It seems to me that everything in the code is char*. What am I missing?

I've looked over this code many times and can't seem to find what is wrong. I'm relatively new to pointers, go easy! Thank you very much for any help!

Answer 1

*(t + i)

t is of type char * .

so t + i means char * + i which means "add the value of i to the pointer to make a new pointer". *(t + i) then dereferences that new pointer, the type of that dereferenced expression will be char . So yes, the compiler is correct. You're trying to pass a char into a function that expects a pointer-to-char .

You simply want

strcpy2(t + i, s);

note: You were also dereferencing s , which would cause the same compile error.

Answer 2

The expression *(t + i) in strcpy2(*(t + i), *s); is a char type because the * dereferences the pointer.

Answer 3

Change these statements

for (i = 0; *t; i++);

strcpy2(*(t + i), *s);

to

for (i = 0; *(t + i ); i++);
strcpy2( t + i, s);

Also it would be better to declare the functions as having return type char * . For example

char * strcat2(char* t, const char* s);