having a formatting problem

[closed account oivD8vqX]

In the program I wrote the 'Y' should line up under the 0059 and the '@' should line up under the 0040.

Function address_view prints correctly for int and char, but not for the double.

I really can't understand why, could someone explain what I'm doing wrong?

  #include<iostream>
#include<limits>
#include<cstring>
#include<iomanip>
#include<stdexcept>

using std::cout;

template<typename T>
void address_view(T* var, size_t size) {
    cout << "The passed size of var is: " << size << '\n'
         << "The address is: " << std::hex << var << '\n';
    size = size > 100 ? 100 : size;
    char view[100];
    std::memcpy(view, var, size);
    for(size_t i = 0; i != size ; ++i) cout << std::setw(4) << std::setfill('0') << std::hex << static_cast<int16_t>(view[i]) << ' '; 
    cout << '\n';
    for(size_t i = 0; i != size ; ++i) cout << std::setw(4) << std::setfill(' ') <<view[i] << ' '; 
    cout << '\n';
}

int main() try {
    int d1{0};
    int d2{'1'};
    int d3{'2'};
    int d4{0x59};
    int d5{0x40};
    char c1{'@'};
    const char* wilson = "wilson";
    double  dbl{100};

    address_view(&d1, sizeof(d1));
    address_view(&d2, sizeof(d2));
    address_view(&d3, sizeof(d3));
    address_view(&d4, sizeof(d4));
    address_view(&d5, sizeof(d4));
    address_view(&c1, sizeof(c1));
    address_view(wilson, std::strlen(wilson));
    address_view(&dbl, sizeof(dbl));

}
catch (std::exception& e) {
    cout << "Caught: " << typeid(e).name() << '\n'
         << "Message: " << e.what()        << '\n';
    return 1;
}

this is my output:

The passed size of var is: 4
The address is: 0x7ffed53169c8
0000 0000 0000 0000 
                
The passed size of var is: 4
The address is: 0x7ffed53169c4
0031 0000 0000 0000 
   1             
The passed size of var is: 4
The address is: 0x7ffed53169c0
0032 0000 0000 0000 
   2             
The passed size of var is: 4
The address is: 0x7ffed53169bc
0059 0000 0000 0000 
   Y             
The passed size of var is: 4
The address is: 0x7ffed53169b8
0040 0000 0000 0000 
   @             
The passed size of var is: 1
The address is: @@
0040 
   @ 
The passed size of var is: 6
The address is: wilson
0077 0069 006c 0073 006f 006e 
   w    i    l    s    o    n 
The passed size of var is: 8
The address is: 0x7ffed53169a0
0000 0000 0000 0000 0000 0000 0059 0040 
                           Y    @ 

[ne555]

you are trying to print a non-printable character ('\0') so nothing is printed
but the filler was already done, to just three spaces instead of four


> The address is: @@
cast `var' to void*, if you try to print a char* it will go on until it finds a null terminator.

[closed account oivD8vqX]

I'm sorry, I left them in, the chars ints, to show that they're printing correctly, my problem is with the double.

It's address was: 0x7ffed53169a0.

This is the output I don't understand.

The passed size of var is: 8
The address is: 0x7ffed53169a0
0000 0000 0000 0000 0000 0000 0059 0040 
                           Y    @ 

My suspicion, although I don't know, is the '@@' isn't an address at all. I've been told that sometimes the compiler optimizes some variables away.

[closed account oivD8vqX]

@ness555-- I'm sorry, I just figured out what you're saying. I think I can fix it now. Thank you.

[closed account oivD8vqX]

for completeness here is the fixed function. Thank you again.

template<typename T>
void address_view(T* var, size_t size) {
    cout << "The passed size of var is: " << size << '\n'
         << "The address is: " << std::hex << (void*)var << '\n';
    size = size > 100 ? 100 : size;
    char view[100];
    std::memcpy(view, var, size);
    for(size_t i = 0; i != size ; ++i){
        cout << std::setw(4) << std::setfill('0') << std::hex << int{view[i]} << ' '; 
    }
    cout << '\n';
    for(size_t i = 0; i != size ; ++i) {
        if (view[i] == 0) cout << "     "; 
        else cout << std::setw(4) << std::setfill(' ') <<view[i] << ' ';
    } 
    cout << '\n';
}

edit: added cast to solve char* address problem.