为什么我们在使用 scanf 函数时需要指定变量的地址而不是变量的名称

Question

Why do we need to specify the address of variable and not the name of variable while using scanf function for eg: scanf("%d",&a); why we use & instead we could have written like this scanf("%d",a); so why we need to use address while using scanf() ?

Answer 1

scanf would implicitly want an address at all cases, because the variable you will have as input must be stored somewhere in the memory, and using &variable_name you are assigning a pointer addressing the variable's location.

In case of arrays/strings, you can go without using & as because the array/string passed is infact a pointer to where the first value/character is stored. (from where you can iterate over it using pointer arithmetic, for eg: +4 for int)

Answer 2

above answers and comments gives you technical reasons but in laymen terms

so your question is basically why scanf cannot take just name of variable instead of the address.

You use scanf() to put something in a variable right?

Now lets suppose you want to deliver a letter , can you deliver it by just giving the name of the person ? NO

You need to put address on it then the postman will reach that address, the postman does not need to know the name of receiver he only needs to know the address.

Your scanf like that postman it needs the address of memory location to deliver that value to the variable you want.