When I run the following C code I get the values:
222222222
312222222
102222222
I was expecting the values:
222222222
31
10
Why does the char number[] defined in my intToStr function remember previous values? I thought once the function call ended all local data was more or less destroyed.
#include <stdio.h>
void intToStr(int n);
int main(void)
{
intToStr(222222222);
intToStr(31);
intToStr(10);
return 0;
}
void intToStr(int n)
{
char number[10];
int l = 0;
if (n < 0)
{
l++;
number[0] = '-';
n *= -1;
}
int nCopy = n;
while (nCopy > 9)
{
nCopy /= 10;
l++;
}
int r;
while (n > 9)
{
r = n % 10;
n /= 10;
number[l--] = r + '0';
}
number[l] = n + '0';
printf("%s\n", number);
}
the array should not remember the old data
For each program, the C standard either:
specifies what the program should do
says that it is not specified what the program should do
It hardly ever says that the program should not do something in particular.
In this case, the standard says that it is not specified what characters should be in the array at the start of the function. They can be anything at all. Characters from the previous call is one particular case of "anything at all".
That's undefined behavior. If only the first 3 character are set, it may print 312222222 or it may print 312???????????????????
The last characters in char number[10] are not initialized, that means the compiler may decide to leave it alone and the old values stay, or something else happens.
Otherwise printf doesn't know where the string end, it keeps printing until it randomly hits a zero.
If there is buffer overrun printf finds a different set of characters in memory (which we are not supposed to be accessing) and the program keeps printing those same characters until it randomly hits a zero and finally stops.
To fix it, simply make sure there is '\0' at the end. You can also add additional check to make sure the length does not exceed the buffer size
Working example:
char number[10];
int l = 0;
if (n < 0)
{
l++;
number[0] = '-';
n *= -1;
}
if (n < 0) return;
int nCopy = n;
while (nCopy > 9)
{
nCopy /= 10;
l++;
}
int len = l;
if (len + 1 > sizeof(number))
return;
number[len + 1] = '\0';
int r;
while (n > 9)
{
r = n % 10;
n /= 10;
number[l--] = r + '0';
}
number[l] = n + '0';
printf("%s\n", number);
I am trying to convert from a popen pass, to a float as the final result. I have tried converting to a char, and then into a float in every possible way I can find, however the output I have seen using printf seems to be wrong every time. I have tried using a tostring function, as well as using a %s like in the printf function that returns the correct function, however it all seems to give me the wrong output as soon as I try to convert the output. Should I be trying a different conversion method?
Here is the code.
FILE * uname;
char os[80];
int lastchar;
char n;
uname = popen("sudo python ./return63.py", "r");
lastchar = fread(os, 1, 80, uname);
os[lastchar] = "\0";
n = toString(("%s", os));
printf("THE DIRECT OUTPUT FROM PY IS %s", os);
printf("THE DIRECT OUTPUT For n IS %c", n);
float ia = n - 0;
long p = ia - 0;
float dd = p - 0;
printf("Your OS is %f", dd);
Output from the PY is 'THE DIRECT OUTPUT FROM PY IS 63.0' , which is the correct value,
output from the n is 'THE DIRECT OUTPUT For n IS �'
output from the dd is 'Your OS is Your OS is 236.000000'
The function tostring was pulled from an answered question about how to get the output from another answered question. I have tried with and without this code.
int toString(char a[]) {
int c, sign, offset, n;
if (a[0] == '-') { // Handle negative integers
sign = -1;
}
if (sign == -1) { // Set starting position to convert
offset = 1;
}
else {
offset = 0;
}
n = 0;
for (c = offset; a[c] != '\0'; c++) {
n = n * 10 + a[c] - '0';
}
if (sign == -1) {
n = -n;
}
return n;
}
toString returns an int, so store an int and output an int.
int n = toString(os); // Also removed the obfuscating '("%s", ..)'
printf("THE DIRECT OUTPUT For n IS %d", n);
Also your toString function has undefined behavior because sign might be read without being initialized.
if (a[0] == '-') { // Handle negative integers
sign = -1;
offset = 1;
}
else {
sign = 1;
offset = 0;
}
You have a potential os buffer overflow and you are not doing the null termination of os correctly:
lastchar = fread(os, 1, sizeof(os) - 1, uname); // Only read one byte less
os[lastchar] = '\0'; // changed from string "\0" to char '\0'
And finally you are not checking the input string for digits, you are accepting every input (also the '.' in "63.0"). You might want to stop at the first non-digit character:
for (c = offset; !isdigit((unsigned char)a[c]); c++) {
I am in the process of writing a decipher algorithm for Vegenere Variant Cipher and ran into some C specific issues(I am not too familiar with C).
I get
"Run-Time Check Failure #2 - Stack around the variable 'ch' was corrupted" error.
If I understand the error right, ch is not available when I try to read/write to it(ch in this case represents a HEX value read from the text file, I have posted the code of the function below).
But, for the life of me, I can't figure out where it happens. I close the file way before the I exit the function(exception is thrown at the time I leave the function).
Can you take a look an let me know where I have it wrong? Thanks in advance.
P.S. I am tagging the question with C++ as well as it should pretty much be the same except, maybe, how we read the file in.
Anyways, my code below:
int getKeyLength(char *cipherTxtF){
int potKeyL = 1;
float maxFreq = 0.00;
int winKL = 1;
for (potKeyL = 1; potKeyL <= 13; potKeyL++)// loop that is going through each key size startig at 1 and ending at 13
{
unsigned char ch;
FILE *cipherTxtFi;
cipherTxtFi = fopen(cipherTxtF, "r");
int fileCharCount = 0;
int freqCounter[256] = { 0 };
int nThCharCount = 0;
while (fscanf(cipherTxtFi, "%02X", &ch) != EOF) {
if (ch != '\n') {
if (fileCharCount % potKeyL == 0){
int asciiInd = (int)ch;
freqCounter[asciiInd] += 1;
nThCharCount++;
}
}
fileCharCount++;
}
fclose(cipherTxtFi);
float frequenciesArray[256] = { 0 };
float sumq_iSq = 0;
int k;
for (k = 0; k < 256; k++){
frequenciesArray[k] = freqCounter[k] / (float)nThCharCount;
}
for (k = 0; k < 256; k++){
sumq_iSq += frequenciesArray[k] * frequenciesArray[k];
printf("%f \n", sumq_iSq);
}
if (maxFreq < sumq_iSq) {
maxFreq = sumq_iSq;
winKL = potKeyL;
}
}
return winKL;
}
You are trying to read an hexadecimal integer with fscanf() (format "%02X", where X means "integer in hex format") and store it into a char.
Unfortuantely fscanf() just receives the address of the char and doesn't know that you've not provided the address of an int. As int is larger than a char, the memory gets corrupted.
A solution could be:
int myhex;
while (fscanf(cipherTxtFi, "%02X", &myhex) != EOF) {
ch = myhex;
...
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int myatoi(const char* string) {
int i = 0;
while (*string) {
i = (i << 3) + (i<<1) + (*string -'0');
string++;
}
return i;
}
void decimal2binary(char *decimal, int *binary) {
decimal = malloc(sizeof(char) * 32);
long int dec = myatoi(decimal);
long int fraction;
long int remainder;
long int factor = 1;
long int fractionfactor = .1;
long int wholenum;
long int bin;
long int onechecker;
wholenum = (int) dec;
fraction = dec - wholenum;
while (wholenum != 0 ) {
remainder = wholenum % 2; // get remainder
bin = bin + remainder * factor; // store the binary as you get remainder
wholenum /= 2; // divide by 2
factor *= 10; // times by 10 so it goes to the next digit
}
long int binaryfrac = 0;
int i;
for (i = 0; i < 10; i++) {
fraction *= 2; // times by two first
onechecker = fraction; // onechecker is for checking if greater than one
binaryfrac += fractionfactor * onechecker; // store into binary as you go
if (onechecker == 1) {
fraction -= onechecker; // if greater than 1 subtract the 1
}
fractionfactor /= 10;
}
bin += binaryfrac;
*binary = bin;
free(decimal);
}
int main(int argc, char **argv) {
char *data;
data = malloc(sizeof(char) * 32);
int datai = 1;
if (argc != 4) {
printf("invalid number of arguments\n");
return 1;
}
if (strcmp(argv[1], "-d")) {
if (strcmp(argv[3], "-b")) {
decimal2binary(argv[2], &datai);
printf("output is : %d" , datai);
} else {
printf("invalid parameter");
}
} else {
printf("invalid parameter");
}
free(data);
return 0;
}
In this problem, myatoi works fine and the decimal2binary algorithm is correct, but every time I run the code it gives my output as 0. I do not know why. Is it a problem with pointers? I already set the address of variable data but the output still doesn't change.
./dec2bin "-d" "23" "-b"
The line:
long int fractionfactor = .1;
will set fractionfactor to 0 because the variable is defined as an integer. Try using a float or double instead.
Similarly,
long int dec = myatoi(decimal);
stores an integer value, so wholenum is unnecessary.
Instead of
i = (i << 3) + (i<<1) + (*string -'0');
the code will be much more readable as
i = i * 10 + (*string - '0');
and, with today's optimizing compilers, both versions will likely generate the same object code. In general, especially when your code isn't working, favor readability over optimization.
fraction *= 2; // times by two first
Comments like this, that simply translate code to English, are unnecessary unless you're using the language in an unusual way. You can assume the reader is familiar with the language; it's far more helpful to explain your reasoning instead.
Another coding tip: instead of writing
if (strcmp(argv[1], "-d")) {
if (strcmp(argv[3], "-b")) {
decimal2binary(argv[2], &datai);
printf("output is : %d" , datai);
} else {
printf("invalid parameter");
}
} else {
printf("invalid parameter");
}
you can refactor the nested if blocks to make them simpler and easier to understand. In general it's a good idea to check for error conditions early, to separate the error-checking from the core processing, and to explain errors as specifically as possible so the user will know how to correct them.
If you do this, it may also be easier to realize that both of the original conditions should be negated:
if (strcmp(argv[1], "-d") != 0) {
printf("Error: first parameter must be -d\n");
else if (strcmp(argv[3], "-b") != 0) {
printf("Error: third parameter must be -b\n");
} else {
decimal2binary(argv[2], &datai);
printf("Output is: %d\n" , datai);
}
void decimal2binary(char *decimal, int *binary) {
decimal = malloc(sizeof(char) * 32);
...
}
The above lines of code allocate a new block of memory to decimal, which will then no longer point to the input data. Then the line
long int dec = myatoi(decimal);
assigns the (random values in the) newly-allocated memory to dec.
So remove the line
decimal = malloc(sizeof(char) * 32);
and you will get the correct answer.
if(!strcmp(argv[3] , "-b"))
if(!strcmp(argv[3] , "-d"))
The result of the string compare function should be negated so that you can proceed. Else it will print invalid parameter. Because the strcmp returns '0' when the string is equal.
In the 'decimal2binary' function you are allocating a new memory block inside the function for the input parameter 'decimal',
decimal = malloc(sizeof(char) * 32);
This would actually overwrite your input parameter data.
I have a code which reads around (10^5) int(s) from stdin and then after performing ## i output them on stdout. I have taken care of the INPUT part by using "setvbuf" & reading lines using "fgets_unlocked()" and then parsing them to get the required int(s).
I have 2 issues which i am not able to come over with:
1.) As i am printing int(s) 5 million on stdout its taking lot of time : IS THERE ANY WAY TO REDUCE THIS( i tried using fwrite() but the o/p prints unprintable characters due to the reason using fread to read into int buffer)
2.) After parsing the input for the int(s) say 'x' i actually find the no of divisors by doing %(mod) for the no in a loop.(See in the code below): Maybe this is also a reason for my code being times out:
Any suggestions on this to improved.
Many thanks
This is actually a problem from http://www.codechef.com/problems/PD13
# include <stdio.h>
# define SIZE 32*1024
char buf[SIZE];
main(void)
{
int i=0,chk =0;
unsigned int j =0 ,div =0;
int a =0,num =0;
char ch;
setvbuf(stdin,(char*)NULL,_IOFBF,0);
scanf("%d",&chk);
while(getchar_unlocked() != '\n');
while((a = fread_unlocked(buf,1,SIZE,stdin)) >0)
{
for(i=0;i<a;i++)
{
if(buf[i] != '\n')
{
num = (buf[i] - '0')+(10*num);
}
else
if(buf[i] == '\n')
{
div = 1;
for(j=2;j<=(num/2);j++)
{
if((num%j) == 0) // Prob 2
{
div +=j;
}
}
num = 0;
printf("%d\n",div); // problem 1
}
}
}
return 0;
}
You can print far faster than printf.
Look into itoa(), or write your own simple function that converts integers to ascii very quickly.
Here's a quick-n-dirty version of itoa that should work fast for your purposes:
char* custom_itoa(int i)
{
static char output[24]; // 64-bit MAX_INT is 20 digits
char* p = &output[23];
for(*p--=0;i/=10;*p--=i%10+0x30);
return ++p;
}
note that this function has some serious built in limits, including:
it doesn't handle negative numbers
it doesn't currently handle numbers greater than 23-characters in decimal form.
it is inherently thread-dangerous. Do not attempt in a multi-threaded environment.
the return value will be corrupted as soon as the function is called again.
I wrote this purely for speed, not for safety or convenience.
Version 2 based on suggestion by #UmNyobe and #wildplasser(see above comments)
The code execution took 0.12 seconds and 3.2 MB of memory on the online judge.
I myself checked with 2*10^5 int(input) in the range from 1 to 5*10^5 and the execution took:
real 0m0.443s
user 0m0.408s
sys 0m0.024s
**Please see if some more optimization can be done.
enter code here
/** Solution for the sum of the proper divisor problem from codechef **/
/** # author dZONE **/
# include <stdio.h>
# include <math.h>
# include <stdlib.h>
# include <error.h>
# define SIZE 200000
inline int readnum(void);
void count(int num);
int pft[]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709};
unsigned long long int sum[SIZE];
int k = 0;
inline int readnum(void)
{
int num = 0;
char ch;
while((ch = getchar_unlocked()) != '\n')
{
if(ch >=48 && ch <=57)
{
num = ch -'0' + 10*num;
}
}
if(num ==0)
{
return -1;
}
return num;
}
void count(int num)
{
unsigned int i = 0;
unsigned long long tmp =0,pfac =1;
int flag = 0;
tmp = num;
sum[k] = 1;
for(i=0;i<127;i++)
{
if((tmp % pft[i]) == 0)
{
flag =1; // For Prime numbers not in pft table
pfac =1;
while(tmp % pft[i] == 0)
{
tmp =tmp /pft[i];
pfac *= pft[i];
}
pfac *= pft[i];
sum[k] *= (pfac-1)/(pft[i]-1);
}
}
if(flag ==0)
{
sum[k] = 1;
++k;
return;
}
if(tmp != 1) // For numbers with some prime factors in the pft table+some prime > 705
{
sum[k] *=((tmp*tmp) -1)/(tmp -1);
}
sum[k] -=num;
++k;
return;
}
int main(void)
{
int i=0,terms =0,num = 0;
setvbuf(stdin,(char*)NULL,_IOFBF,0);
scanf("%d",&terms);
while(getchar_unlocked() != '\n');
while(terms--)
{
num = readnum();
if(num ==1)
{
continue;
}
if(num == -1)
{
perror("\n ERROR\n");
return 0;
}
count(num);
}
i =0;
while(i<k)
{
printf("%lld\n",sum[i]);
++i;
}
return 0;
}
//Prob 2 Is your biggesr issue right now.... You just want to find the number of divisors?
My first suggestion will be to cache your result to some degree... but this requires potentially twice the amount of storage you have at the beginning :/.
What you can do is generate a list of prime numbers before hand (using the sieve algorithm). It will be ideal to know the biggest number N in your list and generate all primes till his square root. Now for each number in your list, you want to find his representation as product of factors, ie
n = a1^p1 * a1^p2 *... *an^pn
Then the sum of divisors will be.
((a1^(p1+1) - 1)/(a1 - 1))*((a2^(p2+1) - 1)/(a2-1))*...*((an^(pn+1) - 1)/(an-1))
To understand you have (for n = 8) 1+ 2 + 4 + 8 = 15 = (16 - 1)/(2 - 1)
It will drastically improve the speed but integer factorization (what you are really doing) is really costly...
Edit:
In your link the maximum is 5000000 so you have at most 700 primes
Simple decomposition algorithm
void primedecomp(int number, const int* primetable, int* primecount,
int pos,int tablelen){
while(pos < tablelen && number % primetable[pos] !=0 )
pos++;
if(pos == tablelen)
return
while(number % primetable[pos] ==0 ){
number = number / primetable[pos];
primecount[pos]++;
}
//number has been modified
//too lazy to write a loop, so recursive call
primedecomp(number,primetable,primecount, pos+1,tablelen);
}
EDIT : rather than counting, compute a^(n+1) using primepow = a; primepow = a*primepow;
It will be much cleaner in C++ or java where you have hashmap. At the end
primecount contains the pi values I was talking about above.
Even if it looks scary, you will create the primetable only once. Now this algorithm
run in worst case in O(tablelen) which is O(square root(Nmax)). your initial
loop ran in O(Nmax).