Is there any particular reason why while taking input C (scanf) prefers to take memory location of variable instead of the name of variable itself ? I know its definition of scanf which mandates above ? But is there any special reason ? Like in c++ if u want to take input you'll write cin>>var; But why not in ? Is it something to keep language faster ?
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9Because C passes arguments by copy, without using pointer it cannot modify an object.ouah– ouah2013-06-22 12:32:04 +00:00Commented Jun 22, 2013 at 12:32
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2 Answers
It's because values to functions in C are passed by value. And, of course, being a compiled language names of variables don't exist at run-time, it's all addresses in memory at that point.
If you pass the variable directly, you just pass the value, which is what printf() needs:
int a = 32;
printf("a=%d\n", a);
but if you did the same with scanf()', it would just get the integer value32, with no idea where it came from. Since the point ofscanf()` is to change the value of the caller's variable, you pass the address of the variable:
int a = 32;
scanf("%d", &a);
Then the code inside scanf() can write a new integer value at the given address, which causes the value to change in a.
4 Comments
Harikesh
I got your point and appreciate your response. But my question is why to take memory location instead of variable name. Like in c++ if u want to take input you'll write cin>>var; But why not in ? Is it something to keep language faster ?
tom
@user2511623 C++ introduced references, which (to put it very crudely) are just syntax sugar for pointers. The underlying principle with references (what
cin>> uses) is exactly the same as with pointers (what scanf uses). C did not have references, so the only option was pointers.Harikesh
Thanks tom. Is it one of the reason that makes c is faster than c++ and others?
tom
@Harikesh No. Modern optimizing compilers are extremely complicated and speculations on what would be faster and why are useless.
