Comment by j on "Could anyone explain about scanf using &?"

& pass es the starting address of the string in memory to the function. C is built on the RAM-Model (Random access machine). So you pass the address of the memory segment. the type defines the size of memory used.

so a char is usually 1 Byte long (not utf8) so what you say is:

hey reserve 20 bytes for me and register the starting address to that memory with my variable name;
scan for input from the user and put the characters type starting at the ADDRESS i've given you.

that's why passing the value of it (no &) does not work. you want the program to write at the address / memory block reserved for your variable.

C and pointers gg have fun ^^

You said pasing the value without &, It doesn't work, right? but in this example it's still work.

#include <stdio.h>
int main()
{
    int a;
    char b[10];
    printf("A: \n");
    scanf("%d", &a);

    printf("B: \n");
    scanf("%s", b);

    printf("Value of A = %d and B = %s\n", a, b);
}

It's still working when data type of scanf is char.

I found this explaination, but still not understand well~ stackoverflow.com/questions/5406935/reading-a-str…

Decode sorry i forgot to mention an array is already just a pointer address.

godbolt.org/z/Ua58pU

that's why you need the & for the integer but not the & for the array. since an array is basically just a memory starting address

Appriciated for your explaination

Search Hashnode