copy string to char**

Question

I do not understand why the following code:

char** receive_message(char** ret)
{
    char* temp = "text";

    strcpy(&ret, temp);
    return ret;
}

gives this error:

warning: passing argument 1 of ‘strcpy’ from incompatible pointer type [enabled by default]
     strcpy(&ret, temp);

I'm trying to copy a message that is generated inside a function to a char* array that is allocated outside the function.

You're confusing `&ret` and `*ret`. But why is ret `char**`, and why do you return it? — Roddy, Nov 29 '15 at 20:51
I need a function that gets me a message with msgget as a string. — Mari, Nov 29 '15 at 20:55
There is not even a call to `getmsg`. Please provide a [mcve]. — too honest for this site, Nov 29 '15 at 21:10
And note that `char *? is a _pointer to char_, not an array. — too honest for this site, Nov 29 '15 at 21:10
That was my minimal example, my getmsg works as intended, I just don't understand pointers. — Mari, Nov 29 '15 at 21:11

Andrej Adamenko · Accepted Answer · 2015-11-30T08:40:44.447

2

Use

strcpy(*ret, temp);

Instead of

strcpy(&ret, temp);

You need to deference char ** instead of taking a reference of it.

edited Nov 30 '15 at 08:40

answered Nov 29 '15 at 20:54

Andrej Adamenko

1,650
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score 1 · Answer 2 · answered Nov 29 '15 at 20:53

1

It wants something of type char* whereas &ret is of type char***.

You could maybe pass it *ret which is of type char*.

answered Nov 29 '15 at 20:53

chasep255

11,745
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score 0 · Answer 3 · answered Nov 29 '15 at 20:57

0

ret is a char** but strcpy expects a char*

You then take the address of &ret making it char***

Either pass ret as char* into receive_message or dereference it when passing it to strcpy (*ret, temp)

answered Nov 29 '15 at 20:57

Josh

73
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score 0 · Answer 4 · answered Nov 29 '15 at 20:57

You are passing strcpy the address of a pointer to a pointer to char... That's 2 levels of indirection too many. You do not need all these indirections, you can simplify receive_message this way:

char *receive_message(char *ret) {
    return strcpy(ret, "text");
}

Not really worth a separate function ;-)

It is useful to pass the address of the pointer to receive_function is you intend for this function to modify the pointer in the calling scope, as would be required if you need to reallocate the buffer, but since you do not pass the size, such reallocation would be hazardous.

copy string to char**

4 Answers4