I have declared a character array of size 1 in C but I have read a string of size 10 yet the given code works absolutely fine.

Question

#include<stdio.h>
int main()
{
    char name[1];
    scanf("%s",name);
    printf("%s",name);
}

I am able to read a string of any size with this declaration

Answer 1

Firstly, This is undefined behaviour. What happens in this case is unpredictable.

However, I will give an extra example of code with an additional variable to show what can go wrong with this approach.

int main()
{
    char name[1];
    int a=1;
    printf ("a = %d\n",a);
    scanf("%s",name);
    printf("%s\n",name);
    printf ("a = %d\n",a);
}

This is your original program with an additional variable, a and it is initialized to 1 and not changed.

Output

a = 1
aaaa
aaaa
a = 6381921

As you can see the value of a was changed by entering a large string.

Answer 2

For those new to C, I want to explain what is actually happening: In C, an array is a consecutive block of memory, represented by a pointer to the first cell of that array (first byte of the memory block).

Here is how it would look for array of char with one cell:

  6   7   8   9
+---+---+---+---+
| X | a | X | X |
+---+---+---+---+
      |__ name = 7

You have a byte of memory allocated somewhere and the variable name holds the address of this byte. It does not hold the length of the array, that information is only used to allocate it and then discarded.

When you call scanf it takes the address in name and starts filling the memory with user input. Since scanf has no information about the size, it will only stop in two cases: 1. End of the input 2. Error writing to memory

This is why it is impossible to tell what will happen, as it depends largely on the memory layout of the running program which in turn depends on the hardware, the OS and the compiler.

It also depends on the length of the input.

In this simplified example, scanf will keep writing to addresses 8, 9 and beyond. Assuming these addresses can be written to by your program, it will replace whatever is in there with user input.

printf will then read the memory starting at 7 (address in name) until it runs into a byte with value 0 and print everything to the screen.

If no other code was using that memory between call to scanf and call to printf the input data will still be there and will be printed correctly to the screen.

If however, there was code using those memory addresses, it could overwrite the input or crush the program because it read wrong data from memory.