Here's the code:
class gg{
public:
char ce[10];
gg(char j = ' ') { memset(ce,j,sizeof(ce));};
};
int main() {
gg x('m');
cout << x.ce << endl;
}
It outputs all those 10 ms,and then a series of weird symbols appear,why does that happen?
problems with memset
Author
Your string is not null terminated and so the stream is printing whatever follows the last element until the first null character. Add a '\0' at the end to mark the end of the string.
perhaps the easiest thing to do is:
char ce[10]; // declare string
ce[0]=0; // set string to ""
Uh, no. He's about to overwrite the 10 chars in ce with 10 'm' characters so making it the empty string before you do that isn't very useful.
You need to do this
memset(ce, j, sizeof(ce) - 1);
ce[sizeof(ce)-1] = '\0';
You'll want to up the array size to 11 if you want to store 10 'm' characters in the array though as well as the null terminator.
Note: this only works since sizeof char is defined as 1.
Note: this only works since sizeof char is defined as 1.
You can solve this by using sizeof(char):
memset(ce, j, sizeof(ce) - sizeof(char));
However, unless I'm very mistaken, sizeof(char) is guaranteed to be 1.
Note: this only works since sizeof char is defined as 1.
You can solve this by using sizeof(char):
memset(ce, j, sizeof(ce) - sizeof(char));
However, unless I'm very mistaken, sizeof(char) is guaranteed to be 1.
Only if the char array is previously initialized to zero, which I believe is not the case for local variables (their value, and, for pointers, the memory they point to, is undefined).
EDIT: never mind that. And, yes, sizeof(char) is defined to be 1 by the C and C++ standard. Don't have the references with me now, though. Of course it's best to get in the habit of using vectors (and strings) if using C++, there's no overhead and it's just simpler and safer to use, not to mention more idiomatic.
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