CODEWITHSUNDEEP
c++
vico
vico

Reputation: 18171

Got garbage while char * putting to cout

Simple program that gets char pointer and puts it to the output.

char * get()
{
    char ss [256];
    sprintf (ss,"%d",1);
    return ss;
}

int _tmain(int argc, _TCHAR* argv[])
{
    char *f = get();
    cout<<f;
    char d[50] ;
    cin>> d;
}

I got only garbage on output. Why?

Upvotes: 1

Views: 371

Answers (3)

Les
Les

Reputation: 10605

When the get() function returns, the local variable goes out of scope. I.e., it's value becomes undefined.

To solve, you can use...

static char ss[255];
// or  
char *ss = (char *)calloc(1,255);
// or with C++
char *ss = new char[255];

or so on...

You decide the trade-off. With a static variable, every call to get() could change the contents of the buffer. But with an approach involving allocation, your caller needs to free the memory and know whether to use free() or delete. Some approach the problem by supplying the buffer to the function when called, like...

void get(char *buf, int limit);
// or
void get(char *buf, int& limitActual);

Then main thing is that when dealing with strings, in C/C++ (even std::string) you are dealing with memory that has to be managed somehow. With string, be very careful with automatic variables.

Upvotes: 2

Mahesh
Mahesh

Reputation: 34625

Function is returning the address of a local variable. The code has undefined behaviour and gives unpredicted results.

ss resides on stack and the function get() returns a pointer to it.

char * get()
{
} // Life time of ss ends here

Use std::string instead.

Upvotes: 7

Tonny
Tonny

Reputation: 671

Your array ss[] only exists within the scope of get().
So the pointer to it that you return from get() is invalid as soon as you leave the function.

Upvotes: 2

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