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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;
Related
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I am trying to research various integer overflow scenarios in C and I was wondering does the C language provide any defenses against numeric errors and are there any additional classes or libraries in the C language that can help with that? Also, can anyone give me an example of code that results in an integer overflow in C?
No, there are no defenses.
This overflows:
#include <limits.h>
#include <stdio.h>
const int a = INT_MAX - 2;
const int b = INT_MAX - 2;
printf("%d + %d = %d\n", a, b, a + b);
When I tested it it printed -6, but anything could happen I guess.
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I watched a tutorial which explain int:
but I just get 1:
Enter a:999999999999999999999999999999999999999999
Enter b: 9999999999999999999999999999999999999999999999999999999
a * b = 1
$ ./a.out
Enter a:87999999999999999999999999999999999999999999999999999999999
Enter b: 89999999999998798774334999999999994378969869869869458639534934578365
product = 1
What's the problem?
The reason for you getting 1 instead of -129542144 is that you multiply
999999999999999999999999999999999999999999 and 89999999999998798774334999999999994378969869869869458639534934578365
instead of 300000 and 200000.
Try the following short program you will get the result in the example, -129542144.
#include <stdio.h>
int main()
{
int a = 300000;
int b = 200000;
printf("%d\n", a*b);
return 0;
}
This is assuming that the datatype is int and that it is at least 32 bits.
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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 just wonder how to find mean of two numbers without using division.
do not use these conditions :
int mean = (a + b) >> 1;
four fundamental arithmetic operations
I think this may be helpful -->
int a,b,i,j;
if (a>b)
{
int temp = a;
a = b;
b = temp;
}
for(i=a,j=b;i<j;i++,j--)
continue;
if(i==j)printf("%d\n", i);
else printf("%lf\n", (double)(i)-0.5);
Add them then multiply by 0.5 , no division involved.
If they're both integers, you can use a right shift:
int median = (a + b) >> 1;
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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.