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I am new to C programming but I stumbled on this code
int print(int nb)
{
if (nb < 0)
{
return (0);
}
printf("%d", nb + print(nb - 1));
nb --;
return (nb);
}
int main(void)
{
print(4);
return (0);
}
I ran the code and it gave me an output of 00246
why is that the output that, looking at it logically, the answer is not suppose to start with a 0
print(4) -> print(3) -> print(2) -> print(1) -> print(0) -> print(-1)
print(-1) stops the recursion returning 0, thus a call to printf() is emitted with 0 + 0, which is 0.
print(0) ends with -1 as value, and a call to printf() with 1 + -1 is emitted, which is 0.
etc.
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Errors:
Test isPower2(0[0x0]) failed.
Gives 1[0x1]. Should be 0[0x0]
Code:
int isPower2(int x) {
int nonNega = (x>>31);
int result = !((x & (x-1)) ^ nonNega);
return result;
}
Your isPower2(0) returns true (1) but 0 is not a power of 2. So the expected result would be false (0).
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What does my program not output anything. I have tried other similar versions of solutions for project euler problem 1. I don't need the answer I would just like to know why their is no output. After I compile with gcc and execute the file it seems like is freezes with no output. I have to ctrl-z to kill the program.
#include <stdio.h>
/* Project Euler Problem 1 */
int main()
{
int sum = 0;
int i = 0;
while (i <= 1000);
{
if (i % 3 == 0 || i % 5 == 0);
{
sum += i;
}
i++;
}
printf("%d\n", sum);
return 0;
}
You are closing the while without doing any instruction by putting a semi-colon:
while (i <= 1000);
You should drop the semicolon.
The same for the if instruction:
if (i % 3 == 0 || i % 5 == 0);
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Please excuse me for asking this, because I know the code I'm gonna give you is wrong. Being a newbie I am not able to find the fault. Please help me correct the question and give a solution as well. Again I'm sorry to bother with this simple problem. Tomorrow is my exm in C so i'm kinda desperate. :(
Q: What will be the output of the program?
First let me show you how I find the code first:
#include<stdio.h>
int funct l(int n){
if (n>3)
return int funct(n-3)));
}
main() {
int n= 10;
printf("%d", funct l (n));
}
Then I thought i'd correct it. Then I cleaned up the code as far as i can. Then the code came to this:
#include<stdio.h>
int funct(int n){
if (n>3){
return funct(n-3);
}
}
main() {
int n= 10;
printf("%d", funct(n));
}
still it doesn't give proper answer (though I don't know what it'll show). It is either 1 or 2 and process returned 1 (0*1) is showing at the last line.
Please help me out!
Your funct function doesn't always return a value. This means that it could return anything. Try this:
int funct(int n) {
if (n > 3)
return funct(n - 3);
return n;
}
Here is the call stack when n = 10
funct(n = 10)
funct(n = 7)
funct(n = 4)
funct(n = 1)
return 1
return 1
return 1
return 1
Here is the call stack when n = 11
funct(n = 11)
funct(n = 8)
funct(n = 5)
funct(n = 2)
return 2
return 2
return 2
return 2
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My professor posted this code for us to decipher it for kicks and giggles. It outputs '12 days of christmas'
#include <stdio.h>
int main(t,_,a)
char *a;
{return!0<t?t<3?main(-79,-13,a+main(-87,1-_,
main(-86, 0, a+1 )+a)):1,t<_?main(t+1, _, a ):3,main ( -94, -27+t, a
)&&t == 2 ?_<13 ?main ( 2, _+1, "%s %d %d\n" ):9:16:t<0?t<-72?main(_,
t,"#n'+,#'/*{}w+/w#cdnr/+,{}r/*de}+,/*{*+,/w{%+,/w#q#n+,/#{l,+,/n{n+\
,/+#n+,/#;#q#n+,/+k#;*+,/'r :'d*'3,}{w+K w'K:'+}e#';dq#'l q#'+d'K#!/\
+k#;q#'r}eKK#}w'r}eKK{nl]'/#;#q#n'){)#}w'){){nl]'/+#n';d}rw' i;# ){n\
l]!/n{n#'; r{#w'r nc{nl]'/#{l,+'K {rw' iK{;[{nl]'/w#q#\
n'wk nw' iwk{KK{nl]!/w{%'l##w#' i; :{nl]'/*{q#'ld;r'}{nlwb!/*de}'c \
;;{nl'-{}rw]'/+,}##'*}#nc,',#nw]'/+kd'+e}+;\
#'rdq#w! nr'/ ') }+}{rl#'{n' ')# }'+}##(!!/")
:t<-50?_==*a ?putchar(a[31]):main(-65,_,a+1):main((*a == '/')+t,_,a\
+1 ):0<t?main ( 2, 2 , "%s"):*a=='/'||main(0,main(-61,*a, "!ek;dc \
i#bK'(q)-[w]*%n+r3#l,{}:\nuwloca-O;m .vpbks,fxntdCeghiry"),a+1);
}
That is exactly how he posted it. I have executed it and it works alright, but it's a jumble.
This is an ancient entry into the IOCCC. It has been reverse engineered.