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What will be the result of the following program?
main()
{
char p[]="String";
int x=0;
if(p=="String")
{
printf("Pass 1");
if(p[sizeof(p)-2]=='g')
printf("Pass 2");
else
printf("Fail 2");
}
else
{
printf("Fail 1");
if(p[sizeof(p)-2]=='g')
printf("Pass 2");
else
printf("Fail 2");
}
}
a) Pass 1, Pass 2
b) Fail 1, Fail 2
c) Pass 1, Fail 2
d) Fail 1, Pass 2
e) syntax error during compilation
- 10 Answers
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Answer / jaroosh
Fail 1 , Pass 2.
Some explanation,
1. Fail 1
first of all, to compare strings in C, you use this strcmp
function, so this WOULD give PASS 1 :
if(strcmp(p,"String") == 0)
but
if(p=="String")
will fail because this line means :
if address of p is the same as address of some temporary
storage for literals, where literal "String" is stored,
which is very rarely true, because storing literals is
compiler specific and is very hard to estimate at runtime.
2. Pass 2
sizeof(p) gives 7, because sizeof(char) is 1 byte, and we
have 7 chars in array storing "String", which are :
[0]S
[1]t
[2]r
[3]i
[4]n
[5]g
[6]\0 (EOS)
now, clearly sizeof(p) - 2 is [5] which is "g"
thats why
if(p[sizeof(p)-2]=='g')
is true.
| Is This Answer Correct ? | 10 Yes | 0 No |
Answer / vikram
b)fail1,fail2
bcoz whenever we compare strings,we use strcmp()
function,hence the condition in if() will not be true,
the control will go into else part and will print fail1,then
size of array p is 6 and sizeof(p)-2 results to 4 and hence
p[4]=='n'which again makes the condition in the if()
false,hence fail2 in else part will be printed.
thnx
| Is This Answer Correct ? | 1 Yes | 1 No |
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