1.
int *a;
*a=5;
printf("%d",*a);
Why this one is not giving any output, end with returning big value?
2.
int *a,b=1;
a=&b;
*a=5;
printf("%d",*a);
Why this one works well and shows Output as 5?
3 Answers
In first code snippet
int *a;
a is a pointer and is not pointing to any valid memory location and you dereference the pointer which will lead to undefined behavior.
The second code snippet is good and valid
You have a pointer to a variable b and dereferecnce it.
a = &b;
*a = 5;
Now the pointer is pointing to address of variable b and you change the contents at that location which is defined
Comments
In first snippet, pointer a is not initialized and statement *a = 5; is writing 5 to an unallocated location. If the location modified by this statement belongs to the program, then it may behave erratically. If it belongs to the operating system the program will most likely crash. The behavior is undefined.
Comments
a is an int*, a pointer to int. In the first code snippet, where do you think a points to? The answer to that is a points to some random location. By using
*a=5;
you dereference the pointer and write 5 into the invalid memory location. This invokes Undefined Behavior and anything can happen. If you are lucky, you might end up having a Segmentation Fault.
In the second code snippet, b is an int, an integer variable. The memory for it is supplied by the OS on the stack. You make a point to the address of variable b. Then, dereferencing the pointer and writing 5 to that location is valid as the memory is allocated for variable b.
a, whereas case 2 does:a=&b.