Memory allocation for a char array

[acker]

Hi,

I am trying to create a fresh copy of char array each time the testmemory() function(see below) is called:

void testmemory ()
{
    char * temp = new char [10];
    cout << &temp <<endl;
}
 
int main ()
{
    testmemory();
    testmemory();
    return 0;
}
 

However by observing the output, the memory address of "temp" remains the same at each function call, which is not my expectation. I am not familiar with C++ so pls point out my mistakes.

For your information, I compiled the code under cygwin and mingw, they all give me the same output.

Rick

[acker]

Oh I already see the problem. &temp is the address of the pointer variable, not the address of the char array. But now the problem is, how do I print out the address of the char array? if i try to use cout<<temp; it'll print out the c string not the address

[gpraghuram]

Oh I already see the problem. &temp is the address of the pointer variable, not the address of the char array. But now the problem is, how do I print out the address of the char array? if i try to use cout<<temp; it'll print out the c string not the address

hi,
In C u can get the address like this

printf("temp is %x\n",temp); // This will print address hold by temp
printf("temp is %s\n",temp); //This will print the value
printf("&temp is %x\n",&temp); //This will print the address of temp 
 

Raghuram

[JosAH]

Statically cast your temp variable to a void* and print that.

kind regards,

Jos

[acker]

Thanks Raghuram and Jos, the solutions given by you works perfectly :)