CODEWITHSUNDEEP
c++strcat
Daaksin
Daaksin

Reputation: 894

Unhandled Exception when converting const char to char

I've been trying to convert a const char to a char for the past 30 minutes. Here's what I've got.

 string atr; 

 getline(cin,atr); // Start off with a string because getline takes nothing else.
 const char *buffA = atr.c_str(); // Create a const char of the string converted to a const char.
 char *buff = ""; // Create a new char to hold the converted result.
 strcat(buff,buffA); // Do the conversion.
 parseargs(buff); // Pass the data on.

However, I get an unhandled exception. I have no idea why. I literally just typed 'try' into the console as my only argument.

Upvotes: 1

Views: 251

Answers (4)

orlp
orlp

Reputation: 117691

Try using C++ instead of C idioms:

std::vector<char> data(atr.begin(), atr.end());
data.push_back('\0');
parseargs(&data[0]);

Upvotes: 8

Tom Tanner
Tom Tanner

Reputation: 9354

This

char *buff = ""; // Create a new char to hold the converted result.

creates a char * that points to (probably read-only) memory of about 1 byte in extent. This:

 strcat(buff,buffA); // Do the conversion.

attempts to overwrite that (probably read-only) memory of 1 or so bytes with an arbitrary string.

The chances are this will promptly crash. If the memory is read only, it will crash immediately. If the memory is not read only it will stomp over random data, resulting in very undefined behaviour.

Why on earth do you want to do that? Does parseArgs actually need a modifiable string? It's parsing arguments, it shouldn't need to change them. If it's really necessary, use a std::vector<char> and pass the address of the first element and hope that all it does is poke the contents of the array, rather than (say) running over the end.

Upvotes: 2

James Kanze
James Kanze

Reputation: 153919

There are two things wrong with your code. First, you initialize a char* with a string literal. This uses a deprecated convertion; the type of a string literal is char const[] (which converts to char const*, not to char*), because any attempt to modify the literal is undefined behavior. The second is that your string literal is only one char long, so even if you could write to it, unless atr was empty, you're writing beyond the end of the buffer.

You don't tell us anything about parseargs. If it doesn't modify it's argument, you can just pass it atr.c_str(), and be done with it. (If it's a legacy function which ignores const, you may have to use a const_cast here.) If it does modify its argument (say because it uses strtok), then you'll have to explicitly push a '\0' onto the end of atr, and then pass it &atr[0]. (Not a particularly clean solution, but if you're not using atr afterwards, it should work.)

Upvotes: 5

bash.d
bash.d

Reputation: 13207

Both your contents of buff and buffA are in read-only memory of the process.
You will actually need to new your buff like

char* buff = new char[32];

This provides memory from the free-store and you can then strcat the string from buffA to buff. You should prefer strncat, though to avoid buffer-overruns and delete your buff eventually.

Upvotes: 4

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