What will happen if '&' is not put in a 'scanf' statement?

When a variable is defined, the compiler allocates memory for that variable.

int i;  // The compiler will allocate sizeof(int) bytes for i

i defined above is not initialized and have indeterminate value.

To write data to that memory location allocated for i, you need to specify the address of the variable. The statement

scanf("%d", &i);

will write an int data by the user to the memory location allocated for i.

If & is not placed before i, then scanf will try to write the input data to the memory location i instead of &i. Since i contains indeterminate value, there are some possibilities that it may contain a value equivalent to the value of a memory address or it may contain a value which is out of range of memory address.

In either case, the program may behave erratically and will lead to undefined behavior. In that case anything could happen.

Beacuse it invokes undefined behavior. The scanf() family of functions expect a pointer to an integer when the "%d" specifier is found. You are passing an integer which can be interpreted as the address of some integer but it's not. This doesn't have a defined behavior in the standard. It will compile indeed (will issue some warning however) but it will certainly work in an unexpected way.

In the code as is, there is yet another problem. The i variable is never initialized so it will have an indeterminate value, yet another reason for Undefined Behavior.

Note that the standard doesn't say anything about what happens when you pass a given type when some other type was expected, it's simply undefined behavior no matter what types you swap. But this particular situation falls under a special consideration because pointers can be converted to integers, though the behavior is only defined if you convert back to a pointer and if the integer type is capable of storing the value correctly. This is why it compiles, but it surely does not work correctly.

You passed data having the wrong type (int* is expected, but int is passed) to scanf(). This will lead to undefined behavior.

Anything can happen for undefined behavior. The program may crash and may not crash.

In a typical environment, I guess the program will crash when some "address" which points to a location which isn't allowed to write into by the operating system is passed to scanf(), and writing to there will have the OS terminate the application program, and it will be observed as a crash.

One thing that the other answers haven't mentioned yet is that on some platforms, sizeof (int) != sizeof (int*). If the arguments are passed in a certain way*, scanf could gobble up part of another variable, or of the return address. Changing the return address could very well lead to a security vulnerability.

* I'm no assembly language expert, so take this with a grain of salt.

I could not understand why the program would crash? Could anyone explain me the reason. Any help appreciated.

Maybe a little more applied:

int i = 123;
scanf ("%d", &i);

With the first command you allocate memory for one integer value and write 123 in this memory block. For this example let's say this memory block has the address 0x0000ffff. With the second command you read your input and scanf writes the input to memory block 0x0000ffff - because you are not accessing (dereferencing) the value of this variable i but it's address.

If you use the command scanf ("%d", i); instead you are writing the input to the memory address 123 (because that's the value stored inside this variable). Obviously that can go terribly wrong and cause a crash.