Clicker Questions - 01-C-Stuff-1
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#include <stdio.h>
#include <stdlib.h>
typedef struct id {
int i;
double d;
} ID;
int main(int argc, char **argv) {
ID x, y;
struct id z;
ID *p;
char s[20], *t;
// Question Here
}
Question 1: Which of the following lines will not compile (answer all that apply)?
A.
"Fred" = s;
B.y->d = 3.14;
C.x = y;
D.p—>i = x.i;
E.s = t;
F.s = argv[1];
G.s[30] = 'J';
H.p = &z;
I.x.i = y.d;
J.s[19] = x.i;
Answer
A. "Fred" = s; - This doesn't compile, because you can't set a string to anything.
B. y->d = 3.14; - This doesn't compile, because y is a struct and not a pointer.
You need to do "y.d"
C. x = y; - This compiles fine. It will copy y to x. I don't like it, but it's legal.
D. p—>i = x.i; - This compiles fine. Now, I'm not saying it's going to run correctly.
In particular, if we haven't initialized p, we'll be lucky to get a
segmentation violation with this. However, it complies fine.
E. s = t; - This doesn't compile. You can't set an array to anything.
F. s = argv[1]; - This doesn't compile. You can't set an array to anything.
G. s[30] = 'J'; - This will compile, but some compilers will give a warning,
since you are acessing s out of bounds. Some compilers won't care.
H. p = &z; - Perfectly legal.
I. x.i = y.d; - This is legal, too. It will convert y.d to an integer.
J. s[19] = x.i; - This is legal. It will convert x.i to a one-byte integer.
Extra Answer
On question J, a student in 2024 ask the following question:I have a question about the last question for today’s clicker. You said that assigning x.i to s[19] will typecast it to a "1 byte integer". I am not really sure what that means because to my understanding anything that is 1 byte is simply a char. How does the compiler know its an integer if it’s one byte?
I'm sorry for the confusion. A char is indeed a one-byte integer. It is either signed, in which case it ranges from -128 to 127, or it is unsigned, in which case it ranges from 0 to 255. Sometimes students get confused because chars represent characters, but those characters are just integers. Their character conversion is a standard called ASCII, which, for example, maps 'a' to 97 and 'A' to 65. They are not stored as characters, but they are stored as one-byte integers, so the string "Aa" is really stored in three one-byte integers: {65,97,0}.
In the clicker question, the variable s is declared as:
char s[20];Which means that 20 bytes are allocated for s, and when you access s[0], that will access the first of the 20 bytes, and when you access s[19], that will access the last of the 20 bytes. When I say:
char s[19] = x.i;I am saying store x.i in the last of these 20 bytes. Since x.i is an integer, it is first converted to a char (one-byte integer) before being stored. So, if x.i = 0x123, then 0x23 will be stored in s[19].
I hope that helps answer your question -- JP