const char to char problem

Question

How do I convert from char to const char? How do I implement type casting without changing the char SendBuf? Here is my code:

.cpp

// main()    
char SendBuf[6] = "Hello";
sendto(..., SendBuf, ...);

// sendto structure
int sendto(
__in  SOCKET s,
__in  const char *buf,
__in  int len,
__in  int flags,
__in  const struct sockaddr *to,
__in  int tolen
);

Error

Error   1   error C2440: '=' : cannot convert from 'const char [6]' to 'char [6]'

Thank you.

That code should not produce that error message. Maybe you left something relevant out. — aschepler, Mar 18 '11 at 18:10
please create the smallest program that you can that still has the problem, and copy and paste that code into your question. Please do not re-type your code, but rather paste the exact text in. In summarizing and retyping your program, you left out an important detail. — Robᵩ, Mar 18 '11 at 20:25
possible duplicate of [Array Assignment](http://stackoverflow.com/questions/5279082/array-assignment) — Ben Voigt, Mar 18 '11 at 21:28
@Adams this is my simplest coding what `sendto` from socket parameter should take. i just need to know the correct assignment for my variable. — Chicko Bueno, Mar 19 '11 at 04:59

score 4 · Answer 1 · answered Mar 18 '11 at 20:30

I don't have a Microsoft compiler handy to test this theory, but I suspect that the OP has code like this:

int main() {
    char SendBuf[6];
    SendBuf = "Hello";
    // sendto(..., SendBuf, ...);
}

I suspect that the error he is seeing, cannot convert from 'const char [6]' to 'char [6]' is occuring on assignment, not initialization nor the sendto call.

Can someone with a Microsoft compiler check compile the above program and confirm the error message?

Yes, repro in Visual C++ 2010. – Ben Voigt Mar 18 '11 at 21:27 — Ben Voigt, Mar 18 '11 at 21:27

activout.se · Accepted Answer · 2011-03-18T18:13:47.693

3

Write the assignment like this:

const char *SendBuf = "Hello";

You don't have to do anything about the function call. (A const char* parameter will take a char* variable as input.)

edited Mar 18 '11 at 18:13

answered Mar 18 '11 at 18:08

activout.se

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The error is in the assignment, not the function call. I'll update. – activout.se Mar 18 '11 at 18:10
what does a `const char` is actually means? – Chicko Bueno Mar 18 '11 at 18:23
@Chicko: In a variable declaration, it means it creates a constant variable (can never be written to). On a pointer, it means the pointer is a read-only view of the memory, but some other part of the program might still change to pointed-to variable. – Ben Voigt Mar 18 '11 at 18:35
@divide: I wouldn't recommend this fix, maybe he needs to write into the buffer before transmitting it. – Ben Voigt Mar 18 '11 at 18:36
@divide: What assignment? The assignment that's in the real code, but not in the question? (Your answer and the question both have *initialization*, not *assignment*) – Ben Voigt Mar 18 '11 at 21:31
Actually my buffer would be a dynamic buffer. Means, each time the user needs to specify buffer data before sending. Not just constant variable all the time. – Chicko Bueno Mar 19 '11 at 04:47
Your question is related with http://stackoverflow.com/questions/6024574/how-to-convert-a-const-char-to-simply-char – Mihai8 Jan 08 '13 at 21:48

Ben Voigt · Answer 3 · 2011-03-18T21:32:38.927

2

Try

char SendBuf[6] = { "Hello" };

Note the braces, which tell the compiler you're initializing an aggregate (in this case an array).

And then, instead of

SendBuf = "Hello";

which is obviously your real code, use

strcpy(SendBuf, "Hello");

or

memcpy(SendBuf, "Hello", 6);

edited Mar 18 '11 at 21:32

answered Mar 18 '11 at 18:33

Ben Voigt

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+1: ignores the misleading title & questions resolves the compile error and actual problem ;) – AJG85 Mar 18 '11 at 19:18
-1: It is legal to initialize `char[]` with a string. The braces don't tell the compiler anything it doesn't already know. – Robᵩ Mar 18 '11 at 20:29
@Rob: Well, there's apparently some compiler (probably some older Visual C++) that has a problem with it... actually just tried assignment, since that's what the compiler message talks about, verified that's the issue, and noted you've guessed that as well. – Ben Voigt Mar 18 '11 at 21:27
@Rob: And `char[]` is an exception to the general rule. Putting braces around array initializers is a good habit for consistency, even if skipping it is specifically allowed for `char[]` (only). – Ben Voigt Mar 18 '11 at 21:29
@Rob: Edited answer to address what is obviously the real code. – Ben Voigt Mar 18 '11 at 21:32

score 1 · Answer 4 · answered Mar 18 '11 at 18:15

1

Is there a reason why you can't pass your string literal into the method? sendTo(...,"Hello",...);

If yes then why not simple declare const char* SendBuf = "Hello"; instead of using the char array as one alternative.

Or perhaps better yet start exploring STL and use std::string SendBuf("Hello"); and pass SendBuf.c_str() into the method.

There is always more than one way to skin a cat just pick the knife that is most comfortable for you.

answered Mar 18 '11 at 18:15

AJG85

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Yuck. `std::string` is not really appropriate for storing network packets. And `c_str` is definitely not, `sendto` doesn't work on the basis of a NUL terminator. – Ben Voigt Mar 18 '11 at 18:37
I wasn't jumping to the conclusion of usage although you're probably right even though OP only asked about conversion and `char` vs `const char` ... the point remains the same use the best tool for the job whatever that may be. – AJG85 Mar 18 '11 at 19:03

const char to char problem

4 Answers4