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Can someone please explain me the concept of [\] here?
#include<stdio.h>
int f(int t[\]){
return t[0\] + t[2\];
}
int main(void){
int i,a[\] = {-2,-1,0,1,2};
i = f(a+2);
printf("%d",i);
return 0;
}
I see no reason for '\' in your C code. Maybe it is some leftover from copying the code since it is consistently proceeding the ] closing brackets.
If you remove '\' from your program it will compile and the function f(a+2)
will give you the sum of third and fifth element in the array a[].
#include<stdio.h>
int f(int t[])
{
return t[0] + t[2];
}
int main(void){
int i,a[] = {-2,-1,0,1,2};
i = f(a+2); // (a+2) -> { 0, 1, 2 }
printf("%d",i);
return 0;
}
Output:
2
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I executed a code in the C language, However I am unable to understand its output.
#include <stdio.h>
int main()
{
int a=5;
int b= ++a + 0!=0;
printf("%d %d",++a, b);
return 0;
}
The output for the above program is
7 1
I am unable to understand why it is so.
Order of operations causes this to be treated as:
int b = (((++a) + 0) != 0);
Therefore:
int b = (6 != 0);
6 isn't 0, so that has a value of true aka 1.
int b = 1;
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This code prints the array elements, but I can't understand how does k[x-1] gives the array elements.
#include<stdio.h>
int main()
{
int x[]={2,4,6,8,10},k=1;
while (k<=5)
{
printf ("%3d",k[x-1]);
k++;
}
return 0;
}
Array indexes start at 0 in C. An array like int x[]={2,4,6,8,10} will have a value x[0]=2 and so forth. Typically, when iterating through an array, a convention like this is used:
for (int i = 0; i < length; i++)
printf("%3d",x[i]);
Since the code you provided begins the indexing at 1, you have to subtract one to fetch the proper element.
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I was trying to explain to my friend something about C coding and he asked me why his code (with "scanf") didn't work.
#include
int main() {
char x=scanf("%c",&x);
printf("%c\n",x);
return 0;
}
and this one yes
#include <stdio.h>
int main()
{
int k;
char x=getchar
printf("%c\n",x);
return 0;
}
When scanf completes, x contains the character that was read. However, that value is immediately overwritten when x is assigned the return value of scanf, which is the number of items successfully matched or EOF in the event of an error.
If you call scanf without assigning the return value to x you should get the expected result.
For example, this should work.
char x;
scanf("%c",&x);
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#include <stdio.h>
int main()
{
int var=0;
for(; var++; printf("%d",var));
printf("%d", var);
}
Please explain to me this C code. How is the output 1?
You might be confused because of the wrong code indentation. Your code is:
for(; var++; printf("%d",var))
;
printf("%d", var);
So you always get the output of the second printf. As var is initialized to 0 and var++ (the for-condition) is always executed, you end up with var==1.
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What does the main problem in my algorithm,I need to find the smallest positive number which divided from 1 to 20 with out divider...
#include <stdio.h>
#include <stdbool.h>
int main(int argc,char* argv[])
{
int num,j=2;
int saveNum=20;
bool flag = false;
while(!flag)
{
num = saveNum;
while(num%j==0 && j<=20)
{
num /= j;
j++;
}
if(j>20)
flag = true;
else
{
saveNum++;
j=1;
}
}
printf("Done");
printf("%d",saveNum);
}
Are you missing a printf to see what your intermediate results are? That might help you get an idea as to what is going on internally.
But I don't really understand what you're trying to solve. Do you want the result: 2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*17*18*19*20?
Because I think that's what you are currently computing. However, you'll overflow before you get there and iterating to that point will take you a while.
If instead you're trying to find the smallest number that is divisible by every number less than or equal to 20, then you may want to revisit your update of num /= j.