int main(int argc, char *argv[])
{
  
  char b[]="Among Us";
  char* string2 = &b[0];
  
  printf("Print address of b\t\t=  %p\n", &b[0]);
  
  printf("Print content of string2\t=  %p\n", string2);
  printf("Print content of string2\t=  %s\n", string2); // why ???
  
  system("PAUSE");  
  return 0;
}

Why does the last printf show us the content of b? Isn't it supposed to show us the address of b but in %s format?

I thought printf("Print content of string2\t= %s\n", *string2); was the correct method of getting the content of b printed out through string2 but apparently it was the wrong way.

CodePudding user response：

Both specifiers %p and %s expect a pointer.

%p will print the value of the pointer.

But %s will use the pointer to print a C string, that is an array of char delimited by '\0'.

If you give *string2, it is not a pointer. It is the value of the object the pointer points to, in your case a char.

Raise the warning level of your compiler to the maximum, and hope that it is smart enough to recognize format strings and to check the arguments.

CodePudding user response：

printf("Print content of string2\t= %s\n", string2);

Will print the entire string even when there is no deference operator before string2, because %s, as defined by the standard, expects a char *, i.e. an address. Refer here

CodePudding user response：

Why does the last printf show us the content of b?

address of first element in an array is an address of the array itself. => you can use string2 as b as well

Variable names don't exist anymore after the compiler runs , program care address of variable

refer this : How are variable names stored in memory in C?

CodePudding user response：

You'd better revisit the memory addresses & pointers chapter of your C book.

Suppose you're running a 32 bits system, which addresses memory using 4 bytes & the memory state of your code is:

// char array "b"
address   value
=======   =====
aaaa:0001 A // hex value 41
aaaa:0002 m // hex value 6d
aaaa:0003 o // hex value 6f
aaaa:0004 n // hex value 6e
aaaa:0005 g
aaaa:0006   // that's a "space" char, hex value 20
aaaa:0007 U
aaaa:0008 s
aaaa:0009 0 // that's a zero, hex value 00

// pointer "string2"
// pointing to "aaaa:0001"
// in reverse order
// depending on endianness of your system
address   value
=======   =====
ffff:0001 01
ffff:0002 00
ffff:0003 aa
ffff:0004 aa

1. Your first printf will print aaaa:0001 as hex; i.e. the memory address of array b.
2. Your second printf will print aaaa:0001 as hex; i.e. the values stored in address ffff:0001 through ffff:0004. From the format specifier %p, the compiler will know that it's going to print a memory address and read the 4 bytes & revert them before printing.
3. Your third printf will print Among Us; From the format specifier %s, the compiler will know that it's going to print a char array & print the chars at the address it was pointed to, until printf reaches to a string terminator which is a 0.