C Questions on Expressions - Revision history

Arjun Suresh at 16:16, 6 July 2014

2014-07-06T16:16:29Z

Arjun Suresh at 10:08, 21 June 2014

2014-06-21T10:08:17Z

Arjun Suresh at 10:05, 21 June 2014

2014-06-21T10:05:26Z

Arjun Suresh at 16:39, 19 June 2014

2014-06-19T16:39:52Z

Arjun Suresh at 16:39, 19 June 2014

2014-06-19T16:39:35Z

Arjun Suresh at 16:39, 19 June 2014

2014-06-19T16:39:09Z

Arjun Suresh at 12:48, 10 June 2014

2014-06-10T12:48:27Z

Arjun Suresh at 12:47, 10 June 2014

2014-06-10T12:47:58Z

Arjun Suresh: Created page with " {'''What will be the output??''' #include #include int main() { int x = 8, y = 1; switch(x--, y..."

2014-06-10T12:43:39Z

Created page with " <quiz display="simple"> {'''What will be the output??''' <syntaxhighlight lang="c"> #include <stdio.h> #include <string.h> int main() { int x = 8, y = 1; switch(x--, y..."

New page

<quiz display="simple">
{'''What will be the output??'''
<syntaxhighlight lang="c">
#include <stdio.h>
#include <string.h>

int main()
{
int x = 8, y = 1;
switch(x--, y++)
{
case 1: x*=8;
case 2: y*= x/=2;
case 3:
case 4: y--;
default: x+=5;
}
printf("%d %d", x, y);
}
</syntaxhighlight>
|type="()"
/}
-64 2
-64 1
+33 55
-33 56
||

</quiz>

@@ Line 92: / Line 92: @@
 -7
 -5
-||comma operator always returns the right operand and it also forms a sequence point (comma in function parameter list is actually just a separator and not comma operator thus does not form a sequence point). Hence first pre increment of i is done and then again pre increment is done. But between the second pre increment and the assignment operation there is no sequence point. So, this produces undefined behaviour.
+||comma operator always returns the right operand and it also forms a sequence point (comma in function parameter list is actually just a separator and not comma operator thus does not form a sequence point). Hence first pre increment of i is done and then again pre increment is done. But between the second pre increment and the assignment operation there is no sequence point (like i = ++i). So, this produces undefined behaviour.
 </quiz>

@@ Line 49: / Line 49: @@
 -24
 ||comma operator always returns the right operand and it also forms a sequence point (comma in function parameter list is actually just a separator and not comma operator thus does not form a sequence point). Hence post increment of i is done before comma operator starts and then pre increment is done and finally i (5) is multiplied by j (6) giving 30.
 </quiz>

@@ Line 36: / Line 36: @@
 int main()
+{
-  int i = 3, j = 6, k;
+   int i = 3, j = 6, k;
-  k = (i++ * j, ++i * j);
+   k = (i++ * j, ++i * j);
-  printf("%d",k);
+   printf("%d",k);
+}

@@ Line 19: / Line 19: @@
     printf("%d %d", x, y);
+}
 </syntaxhighlight>
 |type="()"
@@ Line 32: / Line 33: @@
 #include <stdio.h>
 #include <string.h>
 int main()
+{
-int i = 3, j = 6, k;
+  int i = 3, j = 6, k;
+  k = (i++ * j, ++i * j);
-k = (i++ * j, ++i * j);
+  printf("%d",k);
+}

← Older revision		Revision as of 16:39, 19 June 2014
Line 28:		Line 28:
	\|\| (x--, y++) will return the value of y which is 1, as comma operator always returns the right value. Hence, switch case starts with 1. Before starting the switch case, x is decremented and y incremented also. So, in case 1, x is 7 and y is 2. x is changed to 56 in case 1. Because of no break, all cases are evaluated here. So, in case 2, x becomes 28 and y becomes 56. In case 3 nothing happens. In case 4, y becomes 55 and finally in default case x becomes 33.		\|\| (x--, y++) will return the value of y which is 1, as comma operator always returns the right value. Hence, switch case starts with 1. Before starting the switch case, x is decremented and y incremented also. So, in case 1, x is 7 and y is 2. x is changed to 56 in case 1. Because of no break, all cases are evaluated here. So, in case 2, x becomes 28 and y becomes 56. In case 3 nothing happens. In case 4, y becomes 55 and finally in default case x becomes 33.

		+	{'''What will be the output??'''
		+	<syntaxhighlight lang="c">
		+	#include <stdio.h>
		+	#include <string.h>
		+
		+
		+	int main()
		+	{
		+	int i = 3, j = 6, k;
		+
		+	k = (i++ * j, ++i * j);
		+	printf("%d",k);
		+	}
		+
		+	</syntaxhighlight>
		+	\|type="()"
		+	/}
		+	-Undefined behavior
		+	-Compiler dependent
		+	+30
		+	-24
		+	\|\|comma operator always returns the right operand and it also forms a sequence point (comma in function parameter list is actually just a separator and not comma operator thus does not form a sequence point). Hence post increment of i is done before comma operator starts and then pre increment is done and finally i (5) is multiplied by j (6) giving 30.
		+
	</quiz>		</quiz>

@@ Line 4: / Line 4: @@
 <syntaxhighlight lang="c">
 #include <stdio.h>
 int main()
@@ Line 32: / Line 31: @@
 <syntaxhighlight lang="c">
 #include <stdio.h>
 int main()
@@ Line 53: / Line 51: @@
 <syntaxhighlight lang="c">
 #include <stdio.h>
-#include <string.h>
 int main() {
    int i = 5;
- i = (++i,i++,i) ;
+  i = (++i,i++,i) ;
    printf(" i = %d",i);
@@ Line 75: / Line 73: @@
 <syntaxhighlight lang="c">
 #include <stdio.h>
 int main() {
@@ Line 93: / Line 90: @@
 -5
 ||comma operator always returns the right operand and it also forms a sequence point (comma in function parameter list is actually just a separator and not comma operator thus does not form a sequence point). Hence first pre increment of i is done and then again pre increment is done. But between the second pre increment and the assignment operation there is no sequence point (like i = ++i). So, this produces undefined behaviour.
 </quiz>

@@ Line 26: / Line 26: @@
 +33 55
 -33 56
-|| x--, y++ will return the value of y which is 1, as comma operator always returns the right value. Hence, switch case starts with 1. Before starting the switch case, x is decremented and y incremented also. So, in case 1, x is 7 and y is 2. x is changed to 56 in case 1. Because of no break, all cases are evaluated here. So, in case 2, x becomes 28 and y becomes 56. In case 3 nothing happens. In case 4, y becomes 55 and finally in default case x becomes 33.
+|| (x--, y++) will return the value of y which is 1, as comma operator always returns the right value. Hence, switch case starts with 1. Before starting the switch case, x is decremented and y incremented also. So, in case 1, x is 7 and y is 2. x is changed to 56 in case 1. Because of no break, all cases are evaluated here. So, in case 2, x becomes 28 and y becomes 56. In case 3 nothing happens. In case 4, y becomes 55 and finally in default case x becomes 33.
 </quiz>