Identify the invalid C expression

**Danny.Hoshi** · 02-21-2015

Can you please explain which is invalid and why? Thanks

CAN YOU PLEASE EXPLAIN.
Identify the invalid C expression or choose "all are
valid". Assume all variables are integer and non-zero.
7.
a) a+b-0 b) c+-a%4 c) xm6-24 d) xf3r6+2
e) all are valid

8.
a) xx+yy%zz b) z%(z%z%z)) c) ha+ha+ha d) x23-20.4
e) all are valid

9.
a) 0-0-0 b) mx+bb/c c) xd-wd+2ax d) 32+45%x
e) all are valid

10.
a) ant+int b) one+twx c) three*3 d) four/-4
e) all are valid

11.
a) -a+b-33 b) 3.4%7-ab c) scanf-3/3 d) xaab+-37
e) all are valid

12.
a) max3*3 b) CHAR-5 c) x0x-c+a%a d) a*abc+
e) all are valid

13.
a) 3-x/z+4.3 b) a/b/cd c) x+2-2 d) x²/x-77
e) all are valid

**anduril462** · 02-21-2015

Same as before, we wont do your homework for you. Tell us what you think the answers are and why, and we'll confirm or correct and explain.

**Danny.Hoshi** · 02-26-2015

Can some explain to me how which are invalid and why? I am not asking you to do my homework for me, just explain.

p.s this is not homework, im studying for midterm.

**Matticus** · 02-26-2015

Read post #2 again, and do what it suggests.

If we tell you which are invalid, then we'd be doing the work for you, and that's not how it works here.

**stahta01** · 02-26-2015

Breakdown the problem.
1. Identify the operations, constants and variables in each possible answer
2. Are the variables valid or not
3. Are the operations done on the constants and variables valid or not
4. Are the constants valid or not

Tim S.

**Danny.Hoshi** · 02-26-2015

ok so for 7- all are valid
8- b is not valid because it need bracket at beginning
9 - all are valid
10 - a is not valid cant have int
11 - c is invalid, can't have scanf
12 - d is invalid, you cant have + at the end

are these correct?

**anduril462** · 02-26-2015

Everything looks good, except 11. Note, scanf is not an keyword, like int. It's simply the name of a function in the standard library. You can see it working here:

Code:

$ cat foo.c
int main(void)
{
    int scanf = 42;
    return scanf;
}
$ make foo
gcc -Wall -ggdb3 -pedantic -std=gnu99 -O0 -o foo foo.c -lm -lpthread -lrt
$ ./foo
$ echo $?  # This echos the exit value of the last command
42

Note, I specifically chose not to #include <stdio.h>, which is where scanf is defined. If it's not defined there, it's a perfectly valid, although dumb and confusing, variable/function name.

What's your next guess? Hint, one of them uses an operator with an invalid type, i.e. that type does not work with the operator used.

**Danny.Hoshi** · 02-26-2015

i think its d, you cant have +-.

**whiteflags** · 02-26-2015

Yes you can.

Code:

int main()
{
    int xaab = 0;
    int c = xaab + -37;
    return 0;
}

Compiles with no issues because unary minus just makes a number negative.

**c99tutorial** · 02-28-2015

To explain why something doesn't work you need to reference variable naming rules (what are the rules for choosing C identifiers?) and you should also learn how to add whitespace between tokens to make things clearer, e.g. 3.4%7-ab looks a little hard on the eyes but it is the same as 3.4 % 7 - ab (5 tokens).

It's also good to try out the expressions in some example program as whiteflags has done. For a beginner you should do this for ALL of the expressions, even if you're pretty sure they're right. You will learn to use the output of your compiler to your advantage.

**c99tutorial** · 02-28-2015

On the scanf question, you can also get an integer result with this, because scanf where scanf refers to the function will be decayed to an int in your expression (this is legal but good compilers will issue a warning because its probably not what you want).

Code:

#include <stdio.h>

int main(void)
{
    int res = scanf - 3 / 3;
    printf("%d\n", res); // Print some integer value related to the location of scanf in memory
}

If you declare scanf within the inner block with your own version, then your own version shadows the outer version of scanf.

Code:

#include <stdio.h>

int scanf(const char *fmt, ...); // outer scanf is a function declared in stdio.h

int main(void)
{
    int scanf = 9;            // this inner scanf is now just like any old int
    int res = scanf - 3 / 3;  // 9 - 3 / 3
    printf("%d\n", res);
}

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