What happens when char data type gets an integer value with %d in scanf?

Question

When I run the following code and give the inputs as "1 2 3 4 5", interestingly it gives the result 0 0 0 0 5. I tried several other numbers as inputs but only the last one is printed out as it is. I know I have to use int declaration when I want to get int inputs but I'm wondering why it gives the outputs as 0 except the last one?

int main(){

char a,b,c,d,e;
scanf("%d %d %d %d %d", &a, &b, &c, &d, &e);
printf("%d %d %d %d %d", a, b, c, d, e);

return 0;

}

Example I/O: < 40 41 42 43 44

0 0 0 0 44

< 1 2 3 4 5

0 0 0 0 5

Answer 1

That program is ill formed and causes undefined behaviour. for the %d specifier, scanf expects a pointer to an int. While C does not define exact bit widths for char and int, for simplicity lets say a char is 8 bit and an int is 16 bit.

What probably happens is the following: Your five char variables a,b,c,d,e live in a consecutive block of stack memory like this:

Variable	-	e	d	c	b	a	-
Memory	0x0	0x0	0x0	0x0	0x0	0x0	0x0

When scanf writes the first value (lets say its 1) to variable a, it assumes it is an 16 bit integer and the following happens:

An int with the value 1 (or 0x0001) is written to the memory where &a points. You are probably on a little-endian system, so the 0x0001 is represented as 0x0100 in memory (least significant byte first). You get:

Variable	-	e	d	c	b	a	-
Memory	0x0	0x0	0x0	0x0	0x0	0x1	0x0

Then it writes the next value (say its 2, or 0x0002, represented in memory as 0x0200) to variable b:

Variable	-	e	d	c	b	a	-
Memory	0x0	0x0	0x0	0x0	0x2	0x0	0x0

Note how it overwrites a again because it is writing an int (16 bits) instead of a char (8 bits), but b is only 8 bits wide.

Same happens for the remaining variables.

Answer 2

Let's take a simple example:

char c;
scanf("%d", &c);

scanf with a %d format requires an argument of type int*. It will try to read a value from stdin and, if successful will store an int value in the object to which the argument point.

So if c were of type int, this would be fine.

The standard says that if the type of the argument is incorrect for the format string, the behavior is undefined. That's all it says about it. That means that it might behave exactly the way you want it to (which is in a sense the worst possible outcome, because it means you have a bug that's difficult to diagnose, and might show up at the worst possible moment). Or it could cause your program to crash, or allow it to proceed with invalid data.

So what's actually likely to happen?

scanf, if it successfully reads an int value, will try to store that value in an int object located where c is. If char is 1 byte (which it is by definition) and int is 4 bytes (which is very common), then it will store 4 bytes of data to a 1-byte location. The remaining 3 bytes can clobber whatever is in the memory adjacent to c.

If those 3 bytes aren't allocated to anything, there might not be a visible symptom. If some other declared object is stored there, you could clobber that object. If some implementation-specific data is stored there (say, your function's return address), Bad Things Could Happen.

"Undefined behavior" means that your code is breaking the rules -- but it doesn't mean that the implementation has to do anything about it. The burden is on you, the programmer, to avoid breaking those rules, with or without the compiler's help.

Many compilers will, if invoked with the right options, warn you about this particular issue. Do not ignore those warnings.

You can say a lot about what's likely to happen in the presence of undefined behavior if you know something about the particular implementation you're using. And sometimes that can be useful. More often, though, especially if you're writing new code, your time is better spent fixing your code rather than figuring out all the myriad ways it can go wrong.

Answer 3

What @VladfromMoscow means you trying to scanf() char with (%d) its only for decimal input not char you have 2 ways to fix this:

1. change char to int
2. change %d to %c