I have been trying to make a program in C/C++ that creates files until the process is stopped.
The file names start from 0 and follow an arithmetic sequence.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
long cifre(long x) //returns the number of digits a number has
{
int nr = 0;
while (x != 0)
{
x = x/10;
nr++;
}
return nr;
}
int main()
{
long i=0;
FILE* g;
char* v;
char buffer[1025];
int j=0;
for (j=0;j<1024;j++)
buffer[j] = 'x';
while (1)
{
v = (char*)malloc(cifre(i)+10);
snprintf(v,sizeof(v),"%ld",i);
g = fopen(v,"w");
fprintf(g,"%s",buffer);
free(v);
fclose(g);
i++;
}
return 0;
}
The problem is that the program creates only 1000 files.
The sizeof(v) in the call to sprintf is the size of a char pointer, in your case probably 4, which means that the formatted string will contain at most 3 characters, or the numbers from 0 to 999. To fix this use the same length you used to allocate memory:
size_t len = cifre(i)+10;
v = (char*)malloc(len);
snprintf(v,len,"%ld",i);
snprintf(v,sizeof(v) doesn't make much sense because sizeof(v) returns the size of the pointer (v is char*), not the size of the dynamically allocated array. And so snprintf() is limited to only printing sizeof(v)-1 characters, or 3 digits and the NUL string terminator. 3 digits give you values from 000 to 999, exactly 1000.
Related
I'm learning about arrays in C and I can't figure out why the following is not correct?
#include <cs50.h>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <ctype.h>
int main(void)
{
string plaintext = get_string();
int x = 5;
long long N = strlen(plaintext);
string a = plaintext;
long long c = 0;
int z = x;
for(int i = 0; i < N + (N/x) ; i++)
{
if( i == x)
{
a[c] = 32;
c++;
z = (z + x);
//printf("%c\n", a[c]);
}
a[c] = plaintext[i];
//printf("%c\n", a[c]);
c++;
}
printf("%s\n", a);
}
It's meant to insert spaces into a string of text after every x chars... I know it's not efficient (I reckon I need something called pointers) but why isn't it working? I went through it using a debugger and it seems like my original string is changing as I go... but why?
Assuming string is char * then text and a point to the same string. That explains why your original string changes. What you can do is:
string a= malloc(N+1 + N/x +1);
This allocates space for a new string, into which you copy the original with a space after every x characters. Add 1 for the terminating null character and 1 "to be safe" when x or N are odd.
#include <bits/stdc++.h>
using namespace std;
#define freinput "input.txt","r",stdin
#define freoutput "output.txt","w",stdout
#define mp make_pair
#define fi first
#define sc second
#define ellapse printf("Time : %0.3lf\n",clock()*1.0/CLOCKS_PER_SEC);
typedef long long ll;
typedef unsigned long int uld;
typedef vector<int> vi;
typedef vector<string> vs;
typedef pair<int,int> pii;
string s;
string stringInsertion(int x,string neww){
for(int i = 0;i<s.size();i++){
if(i!=0 && i%x==0){
neww=neww+' '+s[i];
}
else neww+=s[i];
}
return neww;
}
int main(){
cin>>s;
int x = 2;
string neww="";
cout<<stringInsertion(x,neww);
}
just set the x number.hope this help
Okay, let's do something similar first: Print out the string with spaces. Use i to loop through the string. Every time i is evenly divisibly by x, we print a space before we print the character, but not at the beginning:
void print_spaced(const char *s, int x)
{
int i;
for (i = 0; s[i]; i++) {
if (i && i % x == 0) putchar(' ');
putchar(s[i]);
}
putchar('\n');
}
You don't need to determine the length beforehand, because you can stop when you hit the terminating null character. That is, keep going as long as s[i] is not null. (Recall that s[i] is the same as s[i] != '\0' and similarly, i is the same as i != 0.)
Now let's fill a char array with the spaced out string instead of printing it:
int space_out_unsafe(char *res, const char *s, int x)
{
int i, k = 0;
for (i = 0; s[i]; i++) {
if (i && i % x == 0) res[k++] = ' ';
res[k++] = s[i];
}
res[k] = '\0';
return k;
}
This function takes an additional parameter: A char buffer to fill. It has a second index, k, which is the current length of the result buffer. Whenever we printed in the first version, we now append a character to the string:
res[k++] = '#';
Tis overwrites the current end and moves k on one position. We don't write a newline at the end, but we must null-terminate the result.
There is one problem, though: The buffer may overflow; note how I have labelled the function above unsafe. Arrays in C have a fixed size and won't grow automatically when something is appended. It is there fore a good idea to pass the maximum buffer size max to the function and check for overflow before appending:
int space_out(char *res, int max, const char *s, int x)
{
int i, k = 0;
for (i = 0; s[i]; i++) {
if (i && i % x == 0 && k < max - 1) res[k++] = ' ';
if (k < max - 1) res[k++] = s[i];
}
res[k] = '\0';
return k;
}
You can now use this function like this:
char res[20];
space_out(res, sizeof(res), "Doremifasola", 2);
puts(res);
There are other ways to accomplish this. You could allocate the memory dynamically, as Paul suggested. That way, you can cater for the additional space you need, but you also make the caller of the function take care of cleaning up the allocated memory with free. Dynamically allocating memory is something to look into after your first week. :)
Another possibility is to space out the string in place, that modify the contents of the original buffer. You still have to take care to provide the extra space, though. (Usually, in-place midofication is used when the result string is shorter, e.g. when filtering out characters.) You should also process your string from the and as not to overwrite data you need later with spaces. If you feel confident, that's an exercise for next week, too.
trying to convert dec to 32-base, and then print it to a file.
const char digits[] = "0123456789ABCDEFGHIJKLMNOPQRSTUV";
char* baseConverter(int num, int base)
{ char* res;
int i=0;
if (num == 0 || base == 10)
{
snprintf(res,"%03x",num);
return *res;
}
while( num > 0 )
{
*(res+i) = digits[num%base];
num=num/base;
}
return *res;
}
and then at the output code :
sprintf(line, "%03s", baseConverter(i, 32);
but I keep getting that Segmentation fault (core dumped) error at running.
There are several things going on here:
First an uninitialised local pointer has an indeterminate value; it doesn't point anywhere in particular. The NULL pointer doesn't point anywhere either, but at least you can test for a NULL pointer easily. Make a habit of initalising a pointer to make it point to valid memory or to make it explicitly null.
The pointer is supposed to point to a char buffer. The way your function looks like, you must allocate memory for that buffer on the heap with malloc. (You can't use local storage, because that would be invalidated immediately.)
Don't make base 10 a special case. (You're even doing it wrong by printing base 10 numbers as hex.)
Your method of printing is okay, but you print the number backwards. So determine the required klength first and then decrement the position you print at.
Here, you deal with the raw characters. Use res[i] rather than do complicated things with the standard library functions. In particular, don't build strings by concatenating or printing strings to themselves. That's very likely undefined behaviour.
A possible implementation of your function could look like:
int ndigits(int num, int base)
{
int n = 0;
while (num) {
n++;
num /= base;
}
if (n == 0) n++;
return n;
}
char* baseConverter(int num, int base)
{
if (num >= 0 && base > 1 && base <= 36) {
int n = ndigits(num, base);
char *res = malloc(n + 1);
int i = n;
res[n] = '\0';
if (num == 0) res[--i] = '0';
while (num) {
res[--i] = digits[num % base];
num /= base;
}
return res;
}
return NULL;
}
Note how an auxiliary function is used to determine the length of the string. The string is then filled backwards, staring with the null terminator. Also note how invalid cases are handled by returning NULL.
Your calling code must explicitly free the string after using it:
int n = rand() % 100000 + 1;
int m = rand() % 10 + 2;
char *p = baseConverter(n, m);
if (p) printf("%d#%d == %s\n", n, m, p);
free(p);
C has manual memory management and keeping track of allocated stuff is tedious. You can't, for example, call baseConverter from inside printf, because you'd lose the handle to the allocated string.
Another popular variant is to have the calling code allocate the memory and then pas a buffer and its size to the function to fill it. A prototype could then look like this:
void sbase(char buf, size_t buflen, int num, int base);
It would then be called like this:
char buf[33]; // Maximum, when base 2 is printed
sbase(buf, sizeof(buf), 5000, 13);
puts(buf);
Because buf is an automatic variable, no freeing is to be done. (How to implement thins and how to properly enforce that the buffer size isn't exceeded is left as an exercise. :))
The main errors have already been pointed out.
Here is another suggested routine (it doesn't require malloc)
The function sets the value of a pointer to the number of converted digits, to make it easy to print out the required digits.
#include <stdio.h>
/* function takes pointer to array, size of array + number/base
and pointer for number of digits in conversion */
void make32(int *res32, int len, int num, int base, int *rln);
int main()
{
int digits32[20]; // size according to max conversion number in base 32
int len32 = sizeof(digits32)/sizeof(digits32[0]);
int in32, resln, n;
/* convert this number */
in32 = 10000;
/* call function with pointer + size & number/base & ptr to # converted digits*/
make32(digits32, len32, in32, 32, &resln);
/* print out result - reverse order - use number of digits */
for(n = resln; n >= 0; n--) {
printf("%d ", digits32[n]);
}
printf("\n");
return (0);
}
void make32(int *res32, int len, int num, int base, int *rln)
{
int i = 0;
while( num > 0 && i <= len ) {
res32[i] = num % base;
num = num / base;
i++;
}
/* set the number of converted digits */
*rln = i - 1;
}
I want to convert a section of a char array to a double. For example I have:
char in_string[] = "4014.84954";
Say I want to convert the first 40 to a double with value 40.0. My code so far:
#include <stdio.h>
#include <stdlib.h>
int main(int arg) {
char in_string[] = "4014.84954";
int i = 0;
for(i = 0; i <= sizeof(in_string); i++) {
printf("%c\n", in_string[i]);
printf("%f\n", atof(&in_string[i]));
}
}
In each loop atof it converts the char array from the starting pointer I supply all the way to the end of the array. The output is:
4
4014.849540
0
14.849540
1
14.849540
4
4.849540
.
0.849540
8
84954.000000 etc...
How can I convert just a portion of a char array to a double? This must by modular because my real input_string is much more complicated, but I will ensure that the char is a number 0-9.
The following should work assuming:
I will ensure that the char is a number 0-9.
double toDouble(const char* s, int start, int stop) {
unsigned long long int m = 1;
double ret = 0;
for (int i = stop; i >= start; i--) {
ret += (s[i] - '0') * m;
m *= 10;
}
return ret;
}
For example for the string 23487 the function will do this calculations:
ret = 0
ret += 7 * 1
ret += 8 * 10
ret += 4 * 100
ret += 3 * 1000
ret += 2 * 10000
ret = 23487
You can copy the desired amount of the string you want to another char array, null terminate it, and then convert it to a double. EG, if you want 2 digits, copy the 2 digits you want into a char array of length 3, ensuring the 3rd character is the null terminator.
Or if you don't want to make another char array, you can back up the (n+1)th char of the char array, replace it with a null terminator (ie 0x00), call atof, and then replace the null terminator with the backed up value. This will make atof stop parsing where you placed your null terminator.
Just use sscanf. Use the format "ld" and check for return value is one.
What about that, insert NULL at the right position and then revert it back to the original letter? This means you will manipulate the char array but you will revert it back to the original at the end.
You can create a function that will make the work in a temporary string (on the stack) and return the resulting double:
double atofn (char *src, int n) {
char tmp[50]; // big enough to fit any double
strncpy (tmp, src, n);
tmp[n] = 0;
return atof(tmp);
}
How much simpler could it get than sscanf?
#include <assert.h>
#include <stdio.h>
int main(void) {
double foo;
assert(sscanf("4014.84954", "%02lf", &foo) == 1);
printf("Processed the first two bytes of input and got: %lf\n", foo);
assert(sscanf("4014.84954" + 2, "%05lf", &foo) == 1);
printf("Processed the next five bytes of input and got: %lf\n", foo);
assert(sscanf("4014.84954" + 7, "%lf", &foo) == 1);
printf("Processed the rest of the input and got: %lf\n", foo);
return 0;
}
I'm trying to create a char array made of some letters and numbers (the function was way more complex initially but i kept simplifying it to figure out why it doesn't work properly). So i have a char array in which i put 2 chars, and try to add some numbers to it.
For a reason i can't figure out, the numbers do not get added to the array. It might be really stupid but I'm new to C so here's the simplified code. Any help is much appreciated, thanks!
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char some_string[20];
char *make_str() {
some_string[0] = 'a';
some_string[1] = 'x';
int random = 0;
int rand_copy = 0;
random = (rand());
rand_copy = random;
int count = 2;
while ( rand_copy > 0 ) {
rand_copy = rand_copy / 10;
++count;
}
int i=2;
for (i=2; i<count; i++) {
some_string[i] = random%10;
random = random/10;
}
return (some_string);
}
int main(int argc, const char *argv[]) {
printf("the string is: %s\n",make_str());
return 0;
}
You have many problems:
resulting string is not zero-terminated. Add some_string[i] = '\0'; to fix this
character (char) is something like "a letter", but random % 10 produces a number (int) which when converted to character results in control code (ASCII characters 0-9 are control codes). You'd better use some_string[i] = (random % 10) + '0';
you're using fixed length string (20 characters), which may be enough, but it could lead to many problems. If you are a beginner and haven't learn dynamic memory allocation, than that's ok for now. But remember that fixed-length buffers are one of top-10 reasons for buggy C-code. And if you have to use fixed-length buffers (there are legitimate reason for doing this), ALLWAYS check if you are not overrunning the buffer. Use predefined constants for buffer length.
unless the whole point of your excercise is to try converting numbers to strings, use libc function like snprintf for printing anything into a string.
don't use global variable (some_string) and if you do (it's ok for a small example), there is no point in returning this value.
Slightly better version:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define BUF_LENGTH 20
char some_string[BUF_LENGTH];
char *make_str() {
some_string[0] = 'a';
some_string[1] = 'x';
int random = rand();
int rand_copy = random;
int count = 2;
while (rand_copy > 0) {
rand_copy = rand_copy / 10;
++count;
}
int i;
for (i = 2; i < count; i++) {
/* check for buffer overflow. -1 is for terminating zero */
if (i >= BUF_LENGTH - 1) {
printf("error\n");
exit(EXIT_FAILURE);
}
some_string[i] = (random % 10) + '0';
random = random / 10;
}
/* zero-terminate the string */
some_string[i] = '\0';
return some_string;
}
int main(int argc, const char *argv[]) {
printf("the string is: %s\n",make_str());
return 0;
}
I am looking for a (relatively) simple way to parse a random string and extract all of the integers from it and put them into an Array - this differs from some of the other questions which are similar because my strings have no standard format.
Example:
pt112parah salin10n m5:isstupid::42$%&%^*%7first3
I would need to eventually get an array with these contents:
112 10 5 42 7 3
And I would like a method more efficient then going character by character through a string.
Thanks for your help
A quick solution. I'm assuming that there are no numbers that exceed the range of long, and that there are no minus signs to worry about. If those are problems, then you need to do a lot more work analyzing the results of strtol() and you need to detect '-' followed by a digit.
The code does loop over all characters; I don't think you can avoid that. But it does use strtol() to process each sequence of digits (once the first digit is found), and resumes where strtol() left off (and strtol() is kind enough to tell us exactly where it stopped its conversion).
#include <stdlib.h>
#include <stdio.h>
#include <ctype.h>
int main(void)
{
const char data[] = "pt112parah salin10n m5:isstupid::42$%&%^*%7first3";
long results[100];
int nresult = 0;
const char *s = data;
char c;
while ((c = *s++) != '\0')
{
if (isdigit(c))
{
char *end;
results[nresult++] = strtol(s-1, &end, 10);
s = end;
}
}
for (int i = 0; i < nresult; i++)
printf("%d: %ld\n", i, results[i]);
return 0;
}
Output:
0: 112
1: 10
2: 5
3: 42
4: 7
5: 3
More efficient than going through character by character?
Not possible, because you must look at every character to know that it is not an integer.
Now, given that you have to go though the string character by character, I would recommend simply casting each character as an int and checking that:
//string tmp = ""; declared outside of loop.
//pseudocode for inner loop:
int intVal = (int)c;
if(intVal >=48 && intVal <= 57){ //0-9 are 48-57 when char casted to int.
tmp += c;
}
else if(tmp.length > 0){
array[?] = (int)tmp; // ? is where to add the int to the array.
tmp = "";
}
array will contain your solution.
Just because I've been writing Python all day and I want a break. Declaring an array will be tricky. Either you have to run it twice to work out how many numbers you have (and then allocate the array) or just use the numbers one by one as in this example.
NB the ASCII characters for '0' to '9' are 48 to 57 (i.e. consecutive).
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
int main(int argc, char **argv)
{
char *input = "pt112par0ah salin10n m5:isstupid::42$%&%^*%7first3";
int length = strlen(input);
int value = 0;
int i;
bool gotnumber = false;
for (i = 0; i < length; i++)
{
if (input[i] >= '0' && input[i] <= '9')
{
gotnumber = true;
value = value * 10; // shift up a column
value += input[i] - '0'; // casting the char to an int
}
else if (gotnumber) // we hit this the first time we encounter a non-number after we've had numbers
{
printf("Value: %d \n", value);
value = 0;
gotnumber = false;
}
}
return 0;
}
EDIT: the previous verison didn't deal with 0
Another solution is to use the strtok function
/* strtok example */
#include <stdio.h>
#include <string.h>
int main ()
{
char str[] = "pt112parah salin10n m5:isstupid::42$%&%^*%7first3";
char * pch;
printf ("Splitting string \"%s\" into tokens:\n",str);
pch = strtok (str," abcdefghijklmnopqrstuvwxyz:$%&^*");
while (pch != NULL)
{
printf ("%s\n",pch);
pch = strtok (NULL, " abcdefghijklmnopqrstuvwxyz:$%&^*");
}
return 0;
}
Gives:
112
10
5
42
7
3
Perhaps not the best solution for this task, since you need to specify all characters that will be treated as a token. But it is an alternative to the other solutions.
And if you don't mind using C++ instead of C (usually there isn't a good reason why not), then you can reduce your solution to just two lines of code (using AXE parser generator):
vector<int> numbers;
auto number_rule = *(*(axe::r_any() - axe::r_num())
& *axe::r_num() >> axe::e_push_back(numbers));
now test it:
std::string str = "pt112parah salin10n m5:isstupid::42$%&%^*%7first3";
number_rule(str.begin(), str.end());
std::for_each(numbers.begin(), numbers.end(), [](int i) { std::cout << "\ni=" << i; });
and sure enough, you got your numbers back.
And as a bonus, you don't need to change anything when parsing unicode wide strings:
std::wstring str = L"pt112parah salin10n m5:isstupid::42$%&%^*%7first3";
number_rule(str.begin(), str.end());
std::for_each(numbers.begin(), numbers.end(), [](int i) { std::cout << "\ni=" << i; });
and sure enough, you got the same numbers back.
#include <stdio.h>
#include <string.h>
#include <math.h>
int main(void)
{
char *input = "pt112par0ah salin10n m5:isstupid::42$%&%^*%7first3";
char *pos = input;
int integers[strlen(input) / 2]; // The maximum possible number of integers is half the length of the string, due to the smallest number of digits possible per integer being 1 and the smallest number of characters between two different integers also being 1
unsigned int numInts= 0;
while ((pos = strpbrk(pos, "0123456789")) != NULL) // strpbrk() prototype in string.h
{
sscanf(pos, "%u", &(integers[numInts]));
if (integers[numInts] == 0)
pos++;
else
pos += (int) log10(integers[numInts]) + 1; // requires math.h
numInts++;
}
for (int i = 0; i < numInts; i++)
printf("%d ", integers[i]);
return 0;
}
Finding the integers is accomplished via repeated calls to strpbrk() on the offset pointer, with the pointer being offset again by an amount equaling the number of digits in the integer, calculated by finding the base-10 logarithm of the integer and adding 1 (with a special case for when the integer is 0). No need to use abs() on the integer when calculating the logarithm, as you stated the integers will be non-negative. If you wanted to be more space-efficient, you could use unsigned char integers[] rather than int integers[], as you stated the integers will all be <256, but that isn't a necessity.