char s[] = "MAC122";
char *p = s;
printf("%s", p   p[2] - p[1]);

When I run the C code, the output is "C122", but I don't really understand these and - operations with the elements of the string and can't find any online references. Could anyone help me?

CodePudding user response：

Good question. This is caused by the type conversion in C.

Background: Printing char*

Let's start with a simple example:

printf("%s", p);

This will print all the characters starting from the address p, i.e. "MAC122". That's because the string in the memory looks like

And printf keeps printing characters until it finds '\0', i.e. the terminator.

So what if you do this?

printf("%s", p 2);

This will print out "C122", which is exactly your case.

That means the p p[2] - p[1] somehow equal to p 2, right? That's because of the algorithmic operations of between types.

Algorithmic operations between types

First, we know that p[2] is exactly the character C, and p[1] is A.

What happens if we do the operation 'C' - 'A'? That results in the conversion from the char into the ASCII. (It's not really a conversion physically because the computer stores the ASCII value, but logically it helps to understand) Then p[2]-p[1] equal to 'C'-'A' equal to 67 - 65 equal to 2.

Similarly, p p[2] - p[1] equal to p 67 - 65 equal to p 2. Using the former knowledge of how printf works, this explains why you get the "C122" output.

CodePudding user response：

The catch lies here: p p[2] - p[1]. p[2] is 'C', p[1] is 'A'. In C, all characters are represented by their ASCII values and 'C' - 'A' happens to be 2 because 'C' is the second letter after 'A'. So, now we have p 2.

p is a pointer to a char and pointer arithmetic dictates that p be incremented exactly one byte for each increment. Thus, p 2 points to the character 'C' in the string.

printf starts from 'C' and carries on until it finds a '\0' at the end of the string. Which gives you "C122".