For "char list[3][10];" why does all of these work as scanf() %s arguments---&list[i],list[i],&list[i][0]?

Question

Isn't char* the only valid argument type for the %s format specifier used in the format specifier string of scanf()?If so,in my program why each one of these work exactly the same for both scanf()'s and printf()'s %s format specifier:

scanf("%s",&list[i]);
scanf("%s",list[i]);   
scanf("%s",&list[i][0]);

I'll appreciate if you clear the following confusions that arise from this premise:

1) Why is &name[i] working given it is of type char (*)[].Isn't &name[i][0] is the only valid argument as it is of type char* ?

2) Does name[i] decompose/translate into char* when we pass it as an argument for %s?Is it same as passing &name[i][0],which is the address of the first character of each name we enter?Is it why it works?

#include<stdio.h>
#include<string.h>

int main(void)
{

char list[3][10];
int i;
printf("Enter the three names \n");

for(i=0;i<=2;i++)
scanf("%s",&list[i]);  //Why this works?
//scanf("%s",list[i]);   //Does it decompose into char* type?
//scanf("%s",&list[i][0]);

for(i=0;i<=2;i++)
printf("%s\n",list+i);  //All of these printf() work as well
//printf("%s\n",list[i]);
//printf("%s\n",&list[i][0]);

}

Answer 1

list[i] is a char[10], and undergoes array-to-pointer conversion when passed as an argument, so in that place, it is completely equivalent to &list[i][0].

&list[i] is a char(*)[10] and as such an invalid argument for the %s conversion specifier in printf or scanf, invoking undefined behaviour.

However, since the first byte in list[i] has the same address as the array list[i], usually it works⁽¹⁾, since printf and scanf don't know about the types of their arguments and interpret them according to the format string.

⁽¹⁾ It will fail if the representation of a char* is different from that of a char(*)[10]. That would be uncommon, but is possible.