[Mid_Term 12] Fix it (II)

Question

12. There are some mistakes in the code, please fix them by using your compiler, and post your code here. 

(can be answered 2 times)

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[]) {
int a

scanf("%d",a);

if(a%2!=0){
	print("Odd");
}else

print("Even");

	return 0;
}

 

Answer 1

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}

Comments

Case 0: Correct output
Case 1: Correct output

Answer 2

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else
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Comments

Case 0: Correct output
Case 1: Correct output

Answer 3

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Comments

Case 0: Correct output
Case 1: Correct output

Answer 4

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Comments

prog.c: In function 'main':
prog.c:7:9: warning: format '%d' expects argument of type 'int *', but argument 2 has type 'int' [-Wformat=]
 scanf("%d",a);
         ^

---

Case 0: Correct output
Case 1: Correct output

Answer 5

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}

Comments

Case 0: Correct output
Case 1: Correct output

Answer 6

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int main(int argc, char *argv[]) {
int a;

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else
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*** *  **  *  * * **    *  ** * 0;
}

Comments

Case 0: Correct output
Case 1: Correct output

Answer 7

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int ****  argc, char * * {
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}

Comments

prog.c: In function 'main':
prog.c:7:1: error: expected '=', ',', ';', 'asm' or '__attribute__' before 'scanf'
 scanf("%d",a);
 ^~~~~
prog.c:7:12: error: 'a' undeclared (first use in this function)
 scanf("%d",a);
            ^
prog.c:7:12: note: each undeclared identifier is reported only once for each function it appears in
prog.c:10:2: warning: implicit declaration of function 'print' [-Wimplicit-function-declaration]
  print("Odd");
  ^~~~~

Answer 8

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}

Comments

Case 0: Correct output
Case 1: Correct output

Answer 9

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int  *   *  argc, char **** {
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}

Comments

Case 0: Correct output
Case 1: Correct output

Answer 10

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int *  argc, char    ** {
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}

Comments

Case 0: Correct output
Case 1: Correct output