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I want to create a function to take a string like this "2014-02-13T06:20:00"
and to convert to a long long int like 20140213062000.
Has anyone a idea how this can be done?
Here is an algorithm, you will just write the code:
define a long long variable n and initialize it to 0.
for each character c in the string:
if c is a digit, ie greater or equal to '0' and less or equal to '9':
multiply n by 10 and add the value represented by c, store the result in n.
else:
ignore the non digit character
n should have the expected value.
This would convert positive values. If the string has an initial - indicating a negative number, you would test that and negate the number at the end. Note also that overflowing the range of long long int has undefined behavior with this approach.
Your attempt in the comment needs some improvements:
long long string_To_long(const char sl[]) {
long long int n = 0;
for (int i = 0; sl[i] != '\0'; i++) {
char c = sl[i];
if (c >= '0' && c <= '9')
n = n * 10 + (c - '0');
}
return n;
}
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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 no problem in generate random numbers with rand()
I have to create an output file with a string of char e.g. "AQSDJEIOFHDUK"
When the result of rand ()%26 +1 is 1 I've to print in the file "A" when the result is "2" B and so on. I already know a priori how many char will be in the string let's say 50.
I've to do that in C language.
Which function I should use?
Strcat?
Simple set an array with random letters selected via rand()
#define RANDOM_STRING_LENGTH 50
char buf[RANDOM_STRING_LENGTH + 1];
for (size_t i = 0; i < RANDOM_STRING_LENGTH; i++) {
buf[i] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"[rand() %26];
}
buf[RANDOM_STRING_LENGTH] = '\0';
puts(buf);
You can use the ASCII table of characters to generate valid characters. The base for the uppercase alphabet is at 65, so when you generate
int rnd = rand() % 26;
you can say something like
char new_char = 65 + rnd;
or as others have stated, it is easier to read
char new_char = 'A' + rnd;
which should give you one character in the ascii table of characters based off of the result of rand() % 26. Remember, in C an unsigned char is really just a data type that can range from the number 0, to the number 255.
(as stated below, a signed char can range from -126 to 127)
Things like 'a' and 'b' are translated into their ascii equivalent. (i.e. the corresponding number between 0 and 255 that they represent.)
(unless you are using UNICODE of course.)
I hope this helped.
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Good night, what's the best way in C to count the number of possibilities of UNIQUE anagrams in a string with the maximum length of 256 without allows repetitions caused by same letters? The input would be just in uppercase and only alphabet letters are allowed, A-Z. I got stuck in the worst case of the program that got 26! a very large number that overflow even my double. I think I'm very lost here, I'm not so good in C. The program just need to shows the number of possibilities, not the anagram. Like:
LOL = 3
HOUSE = 120
OLD = 6
ABCDEFGHIJKLMNOPQRSTUVWXYZ = 403291461126605635584000000
Thank you guys very much... I tried a lot and failed in every single tried, I'm in distress with it. The way I got more closer of do it was in Pascal, but it also failed in some tests, and I can't use Pascal anyway. I'm using CodeBlocks on Windows that compiles with GCC.
You should calculate factorial of the length of the given string divided by the factorial of the occurrence of every letter.
long double logFactorial (int i) {
return i < 2 ? 0.L : (logFactorial (i-1)+log(long double (i));
}
int countLetter(const char* str, char c) {
int res = 0;
while (str && *str) {
res += *str++ == c;
}
return res;
}
long double numPermutations(const char* str) {
auto res = logFactorial (strlen(str));
for (char c = 'A'; c<='Z'; c++) {
res -= logFactorial (countLetter (str,c));
}
return exp((long double)res);
}
Pay attention!
There are several people here who were correct by saying that factorial of 26 cannot be stored even in 64bit integer.
Therefore, I changed my calculation to the logarithmic of the factorial number and store it in long double which I hope is precise enough (I think that the exp() function is not precise enough)
Nevertheless you cannot use this result as an integer value, unless you find a way to store it in 128bit integer or bigger...
You should test it too if this fits your problem.
There is a faster way to calculate the factorial for this problem by storing the results of 0! up to 26! in an array of the size of [27].
I will leave it to you for asking another question.
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I was wondering if it is possible to index an array by using a binary number instead of a decimal number. For instance, arr[binary].
Yep that's definitely possible. Just prepend your binary number with 0b
int array[] = {1,2,4,6};
printf("%d\n", array[0b0001]); // prints 2
from https://gcc.gnu.org/onlinedocs/gcc/Binary-constants.html binary constants can be written using the 0b syntax
As pointed out, all numbers stored on a computer are binary. Binary is the only thing that can be stored on a computer.
And, C does not support binary syntax. (Or perhaps come C compilers do?)
You could however convert a string from binary like this:
var value = arr[BinaryToInt("1011")];
int BinaryToInt(string s)
{
int value = 0;
int bitValue = 1;
for (int i = s.Length - 1; i >= 0; i--)
{
if (s[i] == '1')
value += bitValue;
bitValue <<= 1;
}
return value;
}
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The program is suppose to convert binary number to decimal form. Only using scanf() and printf() library functions. Takes in a char array from user ---no prompt outputs decimal form, function must be used with parameter (char binaryString[]) after conversion result must be printed out in main. Program does not work don't think I'm converting the binary form to decimal form correctly in function binaryToDecimal since i cant use pow() I'm lost
#include <stdio.h>
#include <math.h>
int binaryToDecimal(char binaryString[]) {
int c, j = 1, decimalNumber = 0;
for (binaryString[c = 0]; binaryString[c] > binaryString[33];
binaryString[++c]) {
while (binaryString[c] != 0) {
remainder = binaryString[c] % 10;
decimalNumber = decimalNumber + remainder * j;
j = j * 2;
binaryString[c] = binaryString[c] / 10;
}
}
return decimalNumber;
}
int binaryToDecimalMain() {
int arraysize = 33;
char binaryString[arraysize];
scanf("%32s", binaryString);
printf("%d",binaryToDecimal(binaryString []);
return 0;
}
I not give you the algorithm because it's seems that you are learning how to program and it is important to you to learn to discover how to solve the problems that are given to you.But I can give you some hints:
use binaryString only to compare with '0' or '1'. Don't try to make any operations like '%' on it.
iterate on the binaryString character by character (no while inside for [this is only for this case, there some algorithm that is necessary to do something like this])
your logic to convert is on the right track
Also you should call your main function main.