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I have an input number that it's value is not always consistent, i mean the value change plus 1 or minus 1 and etc. So, i want to compare it with a const but have a small range value, for example a const int Dist that have value between 14 to 16. Is that possible to implement it on C programming? Please help me.
You can set constant for lower bound, and constant for upper bound and check if the value falls within the range.
Pseudocode:
int const LOWER_BOUND = 14;
int const UPPER_BOUND = 16;
if (input <= UPPER_BOUND && input >= LOWER_BOUND)
... logic here ...
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I'm trying to parse a file that has following data eg:
MAGICNUMBER 400
4 is = 0x34
0 is = 0x30
4
0
0
are different unsigned chars
what i want is those different chars to be converted into
unsigned int x = 400;
when parsing them into my program i want to merge them into one integer i tried bitshifting but it didn't work and i probably did it very wrong and got a very large number probably due misunderstanding of something, what i'm susposed to do to merge those numbers without string tricks and without using std but only using bitshift with a explanation how it works?
Each digit is c - '0'. When you get a new digit, you know that prior ones are one decimal place greater, so you multiply the current number by 10 and add the new digit:
char *s = "400";
int sum = 0;
while(*s >= '0' && *s <= '9') {
sum = 10 * sum + (*s - '0');
s++;
}
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I have to add elements in array.
I have function:
function d = kronDel(j,k)
if j == k
d = 1;
else
d = 0;
end
And i have n=0:31.
I tried this:
x2=j*kronDel(n-2,0);
Why doesnt this work?
I only get x2=0;
You want to make a kronecker delta function but you don't need it.
x2 = j(n==2)
or, if you want to keep the zeros
x2 = j.*(n==2)
If you really want to make the function, just adapt it to:
function d = kronDel(j,k)
d = j==k;
From your comment: "x2=[0,0,j,0,0,0,0...to the 31 all 0s] j=sqrt(-1)"
x2=zeros(1,31);x2(3)=i
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if the range of d is 1<= d <= 10^101 and n is 1<= n <= 200. since the range of double is 2.3E-308 to 1.7E+308. when i take input 11111111111111111111 as d then the value d become 11111111111111111000.000 when i show the value to terminal. that means that it couldn't take the input correctly then how will it give correct value for 10^101. i need know the nth root of d. d will be always in form of p = k^n. that's why i added pow function to know the nth root. but the problem is that the range of p is huge. what i am trying is to solve this problem Power of Cryptography !
int main(){
double d,n;
scanf("%lf%lf", &n, &d))
{
printf("%lf\n", pow(d, 1/n));
}
return 0;
}
A double precision number is not capable of holding all the values between 2.3E-308 to 1.7E+308, it is capable of holding a value between these numbers to a precision of about 15 decimal places.
That means some numbers (such as your example) require more precision than the 8 bytes of data can store.
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#define SRC_ASCLIN_ASCLIN0_RX (*(volatile Ifx_SRC_SRCR*)0xF0038084u)
Here SRC_ASCLIN_ASCLIN0_RX means ASCLIN(Async/Sync serial LIN Comm) Receive Service Request.
I know that the macro is used to point at the address 0xF0038084u. But I want real time examples.
Am working on UART development on Infineon microcontroller.
The macro, when expanded by the preprocessor, cast the integer literal as an address, a pointer to Ifx_SRC_SRCR, then dereferences the pointer so you can get or set the value of the memory stored at that specific address.
So you could write e.g.
Ifx_SRC_SRCR value = SRC_ASCLIN_ASCLIN0_RX;
or
SRC_ASCLIN_ASCLIN0_RX = some_other_value;
It basically equivalent to doing e.g.
int an_integer = 6;
int *pointer_to_an_integer = &an_integer;
*pointer_to_an_integer = 10; // Equivalent to SRC_ASCLIN_ASCLIN0_RX = some_other_value above
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I'm really new to programming so please help me.
Here's my algorithm:
Enter a number
If the number is greater than 26
Subtract 26 from the number
Until the number is less than 26
Then display the final result
How do I write the code for this?
Thank You!
Try this. But next time do put some effort when formulating your question
if (scanf("%d", &n) == 1) {
while (n >= 26) {
n -= 26;
}
printf("%d", n) ;
}