What is the difference between `long long int a=50` and `int a =50LL`?

I have started learning C++ from Josh Lospinoso's C++ Crash Course.

I am unable to wrap my mind around 'types'.

What is the difference between long long int a=50 and int a =50LL ?
To me, they both look like long long int

I tried to run this:

#include<cstdio>
int main() {
    long long int a=50;
    int b=50LL;
    printf("a is %lld and b is %lld",a,b);
}

and got :
> a is 50 and b is 18042367336513586

So there is something wrong with either the format specifier or int b=50LL

On changing the format specifier of b to %d. I get:
> a is 50 and b is 50

So does this mean that b is the default 'int' ? If yes, what is the role of LL then?

On surfing about types, I came across this.
So I tried running this:

#include <iostream>
#include <typeinfo>

int main(){
    long long int a = 50;
    int b = 50LL;
    std::cout << typeid(a).name() << std::endl;
    std::cout << typeid(b).name() << std::endl;

    return 0;
}

which gave
> x<br>
i

So they are indeed of different types (?). But Why?

long long int a=50;

This declares a long long int variable, and initializes it to 50.

int a =50LL;

This declares an int variable, and initializes it to 50. The constant value 50 is lexically specified as a long long integer constant, which gets converted to an int, for the purpose of initializing a (accomplishing absolutely nothing useful, by itself, whatsover).

So the difference is that you end up with either a long long int or an int variable named a. In both cases they are initialized to 50.

The problem is the call to printf, not the initialization.

%lld is the wrong conversion specification for variable b, which has type int. Use %d instead.

• %lld means output a value of type long long int in decimal format (rather than hex or octal).
• %d means output a value of type int in decimal format (rather than hex or octal).

The garbage output is explained by the different sizes of int and long long int.

As you wrote it, printf is going to the address where b is stored, and grabbing 8 bytes, which is (usually) the size of long long int. It should be grabbing only 4 bytes, which is (usually) the size of int.

Here is the correct program:

#include<cstdio>
int main() {
    long long int a = 50LL;
    int b = 50;
    printf("a is %lld and b is %d", a, b);
}

On Microsoft Visual C++, the invalid program actually runs correctly after automatically being fixed by the compiler. Here are the warnings it generated:

1>main.cpp(5,12): warning C4477: 'printf' : format string '%lld' requires an argument of type '__int64', but variadic argument 2 has type 'int'
1>main.cpp(5,12): message : consider using '%d' in the format string
1>main.cpp(5,12): message : consider using '%I32d' in the format string

By the way, the initializer used for b is a poor choice, but perfectly valid. C++ silently performs a narrowing conversion that discards the high-order bits.

int b = 50LL;  // poor form, but legal
int b = 50;    // better form