Binary operation, need help - c

I am about to create a function for a program, this is part of a program and is meant to be a bitmap that holds controls of which memory address is free for use (this has nothing to do with this function to do). The bitmap is bit[64] which holds 8 x 64 bits, the function under is taking a parameter number that is the number of data blocks the function should occupy. In array data_blocks[] should the number to the data block that has bit value 0(free).
Execution of this program gives some strange outputs, and data_blocks[] gives values beyond the length of 512. Can someone please give me a hand? Thanks
#include <stdio.h>
#include <string.h>
void occupyDataBlocks(int number)
{
int ab = number;
char bit[512/8];
int bitNum = 0;
int count;
int data_blocks[ab];
int b;
for(bitNum = 0; bitNum < (sizeof(bit)/sizeof(char)); bitNum++) {
char x = bit[bitNum];
for(count = 0; x != 0; x >>= 1 ) {
if(!(x & 0)) {
data_blocks[count] = count;
}
if(count == number) {
break;
}
count++;
}
if(count == number) {
break;
}
}
if(count == number) {
int a;
for(a = 0; a < 5; a++) {
printf("%d\n", data_blocks[a]);
}
} else {
printf("Not enough data blocks\n");
}
}
int main(void)
{
occupyDataBlocks(3);
return 1;
}

k, where to start ...
1) "sizeof(char)" is most likely 1. So you have a 512-byte array, not a 64-byte array.
2) "bit" array is not initialized.
3) the assignment "char x = bit[bitNum]; " should occur inside the loop.
4) "strlen(bit)" does not do what you think it does. It interprets "bit" as a text string. You probably want do use "sizeof(bit)/sizeof(char)".
5) "(x & 0)" always evaluates to 0. What are you trying to do? If you're trying to test the bit, you want to do "!(x & 1)".
6) "int data_blocks[number]": does this even compile? You can't allocate a local array like that if its size is not known at compile time.
7) if(count == number) {
break;
}
only breaks you out of the inner loop. The outer loop continues on uninterrupted.
8) Do you really want to reset "count" to 0 every iteration of the outer loop? Do you want the code to find 3 free locations somewhere in the array, or 3 free locations in a single byte?

Related

last number in a function array

I want to write a function where I have a given array and number N. The last occurrence of this number I want to return address. If said number cannot be found I want to use a NULL-pointer
Start of the code I've made:
int main(void) {
int n = 3;
int ary[6] = { 1,3,7,8,3,9 };
for (int i = 0; i <= 6; i++) {
if (ary[i] == 3) {
printf("%u\n", ary[i]);
}
}
return 0;
}
result in command prompt:
3
3
The biggest trouble I'm having is:
it prints all occurrences, but not the last occurrence as I want
I haven't used pointers much, so I don't understand how to use the NULL-pointer
I see many minor problems in your program:
If you want to make a function, make a function so your parameters and return types are explicit, instead of coding directly in the main.
C arrays, like in most languages, start the indexing at 0 so if there are N element the first has index 0, then the second has 1, etc... So the very last element (the Nth) has index N-1, so in your for loops, always have condition "i < size", not "i <= size" or ( "i <= size-1" if y'r a weirdo)
If you want to act only on the last occurence of something, don't act on every. Just save every new occurence to the same variable and then, when you're sure it was the last, act on it.
A final version of the function you describe would be:
int* lastOccurence(int n, int* arr, int size){
int* pos = NULL;
for(int i = 0; i < size; i++){
if(arr[i] == n){
pos = &arr[i]; //Should be equal to arr + i*sizeof(int)
}
}
return pos;
}
int main(void){
int n = 3;
int ary[6] = { 1,3,7,8,3,9 };
printf("%p\n", lastOccurence(3, ary, 6);
return 0;
}
Then I'll add that the NULL pointer is just 0, I mean there is literally the line "#define NULL 0" inside the runtime headers. It is just a convention that the memory address 0 doesn't exist and we use NULL instead of 0 for clarity, but it's exactly the same.
Bugs:
i <= 6 accesses the array out of bounds, change to i < 6.
printf("%u\n", ary[i]); prints the value, not the index.
You don't actually compare the value against n but against a hard-coded 3.
I think that you are looking for something like this:
#include <stdio.h>
int main(void)
{
int n = 3;
int ary[6] = { 1,3,7,8,3,9 };
int* last_index = NULL;
for (int i = 0; i < 6; i++) {
if (ary[i] == n) {
last_index = &ary[i];
}
}
if(last_index == NULL) {
printf("Number not found\n");
}
else {
printf("Last index: %d\n", (int)(last_index - ary));
}
return 0;
}
The pointer last_index points at the last found item, if any. By subtracting the array's base address last_index - ary we do pointer arithmetic and get the array item.
The cast to int is necessary to avoid a quirk where subtracting pointers in C actually gives the result in a large integer type called ptrdiff_t - beginners need not worry about that one, so just cast.
First of all, you will read from out of array range, since your array last element is 5, and you read up to 6, which can lead in segmentation faults. #Ludin is right saying that you should change
for (int i = 0; i <= 6; i++) // reads from 0 to 6 range! It is roughly equal to for (int i = 0; i == 6; i++)
to:
for (int i = 0; i < 6; i++) // reads from 0 to 5
The last occurrence of this number I want to return as address.
You are printing only value of 3, not address. To do so, you need to use & operator.
If said number cannot be found I want to use a NULL-pointer
I don't understand, where do you want to return nullpointer? Main function can't return nullpointer, it is contradictory to its definition. To do so, you need to place it in separate function, and then return NULL.
If you want to return last occurence, then I would iterate from the end of this array:
for (int i = 5; i > -1; i--) {
if (ary[i] == 3) {
printf("place in array: %u\n", i); // to print iterator
printf("place in memory: %p\n", &(ary[i])); // to print pointer
break; // if you want to print only last occurence in array and don't read ruther
}
else if (i == 0) {
printf("None occurences found");
}
}
If you want to return an address you need yo use a function instead of writing code in main
As you want to return the address of the last occurence, you should iterate the array from last element towards the first element instead of iterating from first towards last elements.
Below are 2 different implementations of such a function.
#include <stdio.h>
#include <assert.h>
int* f(int n, size_t sz, int a[])
{
assert(sz > 0 && a != NULL);
// Iterate the array from last element towards first element
int* p = a + sz;
do
{
--p;
if (*p == n) return p;
} while(p != a);
return NULL;
}
int* g(int n, size_t sz, int a[])
{
assert(sz > 0 && a != NULL);
// Iterate the array from last element towards first element
size_t i = sz;
do
{
--i;
if (a[i] == n) return &a[i];
} while (i > 0);
return NULL;
}
int main(void)
{
int n = 3;
int ary[] = { 1,3,7,8,3,9 };
size_t elements = sizeof ary / sizeof ary[0];
int* p;
p = g(n, elements, ary); // or p = f(n, elements, ary);
if (p != NULL)
{
printf("Found at address %p - value %d\n", (void*)p, *p);
}
else
{
printf("Not found. The function returned %p\n", (void*)p);
}
return 0;
}
Working on the specified requirements in your question (i.e. a function that searches for the number and returns the address of its last occurrence, or NULL), the code below gives one way of fulfilling those. The comments included are intended to be self-explanatory.
#include <stdio.h>
// Note that an array, passed as an argument, is converted to a pointer (to the
// first element). We can change this in our function, because that pointer is
// passed BY VALUE (i.e. it's a copy), so it won't change the original
int* FindLast(int* arr, size_t length, int find)
{
int* answer = NULL; // The result pointer: set to NULL to start off with
for (size_t i = 0; i < length; ++i) { // Note the use of < rather than <=
if (*arr == find) {
answer = arr; // Found, so set our pointer to the ADDRESS of this element
// Note that, if multiple occurrences exist, the LAST one will be the answer
}
++arr; // Move on to the next element's address
}
return answer;
}
int main(void)
{
int num = 3; // Number to find
int ary[6] = { 1,3,7,8,3,9 }; // array to search
size_t arrlen = sizeof(ary) / sizeof(ary[0]); // Classic way to get length of an array
int* result = FindLast(ary, arrlen, num); // Call the function!
if (result == NULL) { // No match was found ...
printf("No match was found in the array!\n");
}
else {
printf("The address of the last match found is %p.\n", (void*)result); // Show the address
printf("The element at that address is: %d\n", *result); // Just for a verification/check!
}
return 0;
}
Lots of answers so far. All very good answers, too, so I won't repeat the same commentary about array bounds, etc.
I will, however, take a different approach and state, "I want to use a NULL-pointer" is a silly prerequisite for this task serving only to muddle and complicate a very simple problem. "I want to use ..." is chopping off your nose to spite your face.
The KISS principle is to "Keep It Simple, St....!!" Those who will read/modify your code will appreciate your efforts far more than admiring you for making wrong decisions that makes their day worse.
Arrays are easy to conceive of in terms of indexing to reach each element. If you want to train in the use of pointers and NULL pointers, I suggest you explore "linked lists" and/or "binary trees". Those data structures are founded on the utility of pointers.
int main( void ) {
const int n = 3, ary[] = { 1, 3, 7, 8, 3, 9 };
size_t sz = sizeof ary/sizeof ary[0];
// search for the LAST match by starting at the end, not the beginning.
while( sz-- )
if( ary[ sz ] == n ) {
printf( "ary[ %sz ] = %d\n", sz, n );
return 0;
}
puts( "not found" );
return 1; // failed to find it.
}
Consider that the array to be searched is many megabytes. To find the LAST match, it makes sense to start at the tail, not the head of the array.
Simple...

Compilation error - Array index is past the end of array blah.groupsToTrace[LENGTH_OF_A_MACRO] = ToTrace[number];

#include <stdio.h>
#define LENGTH_OF_A_MACRO 3
struct blahS
{
unsigned int groupsToTrace[LENGTH_OF_A_MACRO];
} blahS;
int main()
{
//Value hardcoded to 7 just for testing purpose. Otherwise value is assigned from another function
signed int trace = 7; //trace reads value of range [0-7] from a function
unsigned int ToTrace[LENGTH_OF_A_MACRO];
unsigned int number = 0;
unsigned int noOfGroups = 100;
if (trace != 0)
{
if ((trace == 1)) //b'001
{
ToTrace[number] = noOfGroups / 8;
number++;
}
if ((trace == 4)) //b'100
{
ToTrace[number] = noOfGroups / 2;
number++;
}
if ((trace == 7)) //b'111
{
ToTrace[number] = noOfGroups * 7 / 8;
number++;
}
}
struct blahS blah;
blah.groupsToTrace[LENGTH_OF_A_MACRO] = ToTrace[number]; //Compilation-error
return 0;
}
Basically this is an if-loop which checks and decides groupToTrace based on bit-mapping for a trace value allocated. At the last line i got a compilation error saying - Array index 3 is past the end of the array. I am assigning the calculated groupToTrace values to newPointer_p->groupsToTrace[3] where groupsToTrace[3] is stored in a struct
Question is I got a compilation error as mentioned above with out of bounds access to the array as i understand. But i don't understand where is the mistake.
Compiler version gcc.x86_64 4.8.5-39.el7 #GSS-RHEL7
Any clues or hints highly appreciated. Thanks in Advance!
You seem to misunderstand what this line is doing:
blah.groupsToTrace[LENGTH_OF_A_MACRO] = ToTrace[number];
This is not copying the entire contents of one array to another. It is copying index number of ToTrace to index LENGTH_OF_A_MACRO of blah.groupsToTrace. Both of these indices have the value 3 which is out of bounds for both arrays, as an array of size 3 has indices 0, 1, and 2.
You either need a loop to copy the elements:
int i;
for (i=0; i<number; i++) {
blah.groupsToTrace[i] = ToTrace[i];
}
Or you could use memcpy:
memcpy(blah.groupsToTrace, ToTrace, sizeof(blah.groupsToTrace));

how to initialize array of unknown size in c

I am doing a homework assignment for an intro to programming class in c.
I need to write a program that looks at an int array of unknown size (we are given a initializer list as the test case to use), and determine all the duplicates in the array.
To make sure that an element that was already found to be a duplicate doesn't get tested, I want to use a parallel array to the original that would hold the numbers of all the elements that were duplicates.
I need this array to be the same size as the original array, which of course we don't really know till the initializer list is given to us.
I tried using sizeof() to achieve this, but visual studio says that is an error due to the variable size (const int size = sizeof(array1);) not being constant. Am I not using sizeof correctly? Or is this logic flawed?
Perhaps there is another way to approach this, but I have yet to come up with one.
Here is the code included below, hope the comments don't make it too hard to read.
// Dean Davis
// Cs 1325
// Dr. Paulk
// Duplicates hw
#include <stdio.h>
int main()
{
int array1[] = { 0,0,0,0,123,124,125,3000,3000,82,876,986,345,1990,2367,98,2,444,993,635,283,544, 923,18,543,777,234,549,864,39,97,986,986,1,2999,473,776,9,23,397,15,822,1927,1438,1937,1956,7, 29,- 1 };
const int size = sizeof(array1);
int holdelements[size];
int a = 0; // counter for the loop to initialize the hold elements array
int b = 0; // counter used to move through array1 and be the element number of the element being tested
int c = 0; // counter used to move through holdelements and check to see if the element b has already been tested or found as duplicates
int d = 0; // counter used to move through array1 and check to see if there are any duplicates
int e = 0; // counter used to hold place in hold element at the next element where a new element number would go. sorry if that makes no sense
int flag = 0; // used as a boolian to make sure then large while loop ends when we reach a negative one value.
int flag2 = 0; // used as a boolian to stop the second while loop from being infinite. stops the loop when the end of hold elements has been reached
int flag3 = 0; // used to close the third while loop; is a boolian
int numberofduplicates=0;// keeps track of the number of duplicates found
for (a; a < size; a++)
{
if (a == (size - 1))
holdelements[a] = -1;
else
holdelements[a] = -2;
}
while (!flag)
{
flag2 = 0;
flag3 = 0;
if (array1[b] == -1)
flag = 1;
else
{
while ((!flag) && (!flag2))
{
if (holdelements[c] == -1)
flag2 = 1;
else if (array1[b] == holdelements[c])
{
b++;
c = 0;
if (array1[b] == -1)
flag = 1;
}
}
while (!flag3)
{
if (array1[d] == -1)
flag3 = 1;
else if (array1[b] == array1[d] && b != d)
{
printf("Duplicate of %d, index %d, was found at index %d.\n", array1[b], b, d);
holdelements[e] = d;
d++;
e++;
numberofduplicates++;
}
}
}
b++;
}
printf("Total Duplicates Found: %d\n", numberofduplicates);
return 0;
}
redo to the following:
const int size = sizeof(array1)/sizeof(int);

Segmentation fault (core dumped) error, in a C search function

I'm trying to write a C program to take an array of discrete positive integers and find the length of the longest increasing subsequence.
'int* a' is the array of randomly generated integers, which is of length 'int b'
call:
lis_n = answer(seq, seq_size);
function:
int answer(int* a, int b) {
if (a == NULL) {return -1;}
int i = 0;
int j = 0;
int k = 0;
//instantiate max and set it to 0
int max = 0;
//make an array storing all included numbers
int included[b];
memset(included, 0, b*sizeof(int));
//create a pointer to the index in included[] with the largest value
int indexMax = 0;
//create a pointer to the index in a[]
int indexArray = 0;
//index of a[] for max included
int maxToA = 0;
//set the first included number to the first element in a[]
included[indexMax] = a[indexArray];
//loop until break
while (1) {
if (a[indexArray] > included[indexMax]/*digit greater than last included*/) {
//include the digit
included[indexMax+1] = a[indexArray];
//increment current max pointer
indexMax++;
}
j = b - 1;
while (indexArray >= j/*pointer is at end"*/) {
if (j == (b - 1)) {
if ((indexMax+1) > max/*total is greater than current max*/) {
max = indexMax + 1;
}
}
if (a[b-1] == included[0]/*last element is in included[0], stop*/) {
return max;
} else {
//max included is set to zero
included[indexMax] = 0;
//max included pointer decreased
indexMax--;
//set array pointer to new max included
for (k=0;k<(b-1);k++) {
if (a[k] == included[indexMax]) {
indexArray = k;
}
}
//increment array pointer
indexArray++;
j--;
}
}
indexArray++;
printf("(");
for (i=0;i<b;i++) {
printf("%d,",included[i]);
}
printf(")");
}
}
I'm receiving 'Segmentation fault (core dumped)' in the terminal upon running.
Any help would be awesome.
You have declared
int indexMax = 0;
And here you use it as an array index
incuded[indexMax] = 0;
You increment and decrement it
indexMax++;
...
indexMax--;
You check its range but you don't limit it, you alter the value you compare it with
if ((indexMax+1) > max/*total is greater than current max*/) {
max = indexMax + 1;
}
You never check indexMax against b or with 0
int included[b];
So you are almost guaranteed to exceed the bounds of included[].
Some general points of advice. Make your function and variable names meaningful. Avoid making a premature exit from a function wherever possible. Avoid while(1) wherever possible. And never make assumptions about array sizes (including C "strings"). It might seem hard work putting in the overhead, but there is a payoff. The payoff is not just about catching unexpected errors, it makes you think about the code you are writing as you do it.
I've done something like this for homework before. I got help from:
https://codereview.stackexchange.com/questions/30491/maximum-subarray-problem-iterative-on-algorithm
Make sure you are not trying to index past the size of your array. What I would do would be to find out the size of array a[] (which looks like it is b) and subtract 1. Make sure you are not trying to access past the size of the array.

UVA's 3n+1 wrong answer although the test cases are correct . . .?

UVA problem 100 - The 3n + 1 problem
I have tried all the test cases and no problems are found.
The test cases I checked:
1 10 20
100 200 125
201 210 89
900 1000 174
1000 900 174
999999 999990 259
But why I get wrong answer all the time?
here is my code:
#include "stdio.h"
unsigned long int cycle = 0, final = 0;
unsigned long int calculate(unsigned long int n)
{
if (n == 1)
{
return cycle + 1;
}
else
{
if (n % 2 == 0)
{
n = n / 2;
cycle = cycle + 1;
calculate(n);
}
else
{
n = 3 * n;
n = n + 1;
cycle = cycle+1;
calculate(n);
}
}
}
int main()
{
unsigned long int i = 0, j = 0, loop = 0;
while(scanf("%ld %ld", &i, &j) != EOF)
{
if (i > j)
{
unsigned long int t = i;
i = j;
j = t;
}
for (loop = i; loop <= j; loop++)
{
cycle = 0;
cycle = calculate(loop);
if(cycle > final)
{
final = cycle;
}
}
printf("%ld %ld %ld\n", i, j, final);
final = 0;
}
return 0;
}
The clue is that you receive i, j but it does not say that i < j for all the cases, check for that condition in your code and remember to always print in order:
<i>[space]<j>[space]<count>
If the input is "out of order" you swap the numbers even in the output, when it is clearly stated you should keep the input order.
Don't see how you're test cases actually ever worked; your recursive cases never return anything.
Here's a one liner just for reference
int three_n_plus_1(int n)
{
return n == 1 ? 1 : three_n_plus_1((n % 2 == 0) ? (n/2) : (3*n+1))+1;
}
Not quite sure how your code would work as you toast "cycle" right after calculating it because 'calculate' doesn't have explicit return values for many of its cases ( you should of had compiler warnings to that effect). if you didn't do cycle= of the cycle=calculate( then it might work?
and tying it all together :-
int three_n_plus_1(int n)
{
return n == 1 ? 1 : three_n_plus_1((n % 2 == 0) ? (n/2) : (3*n+1))+1;
}
int max_int(int a, int b) { return (a > b) ? a : b; }
int min_int(int a, int b) { return (a < b) ? a : b; }
int main(int argc, char* argv[])
{
int i,j;
while(scanf("%d %d",&i, &j) == 2)
{
int value, largest_cycle = 0, last = max_int(i,j);
for(value = min_int(i,j); value <= last; value++) largest_cycle = max_int(largest_cycle, three_n_plus_1(value));
printf("%d %d %d\r\n",i, j, largest_cycle);
}
}
Part 1
This is the hailstone sequence, right? You're trying to determine the length of the hailstone sequence starting from a given N. You know, you really should take out that ugly global variable. It's trivial to calculate it recursively:
long int hailstone_sequence_length(long int n)
{
if (n == 1) {
return 1;
} else if (n % 2 == 0) {
return hailstone_sequence_length(n / 2) + 1;
} else {
return hailstone_sequence_length(3*n + 1) + 1;
}
}
Notice how the cycle variable is gone. It is unnecessary, because each call just has to add 1 to the value computed by the recursive call. The recursion bottoms out at 1, and so we count that as 1. All other recursive steps add 1 to that, and so at the end we are left with the sequence length.
Careful: this approach requires a stack depth proportional to the input n.
I dropped the use of unsigned because it's an inappropriate type for doing most math. When you subtract 1 from (unsigned long) 0, you get a large positive number that is one less than a power of two. This is not a sane behavior in most situations (but exactly the right one in a few).
Now let's discuss where you went wrong. Your original code attempts to measure the hailstone sequence length by modifying a global counter called cycle. However, the main function expects calculate to return a value: you have cycle = calculate(...).
The problem is that two of your cases do not return anything! It is undefined behavior to extract a return value from a function that didn't return anything.
The (n == 1) case does return something but it also has a bug: it fails to increment cycle; it just returns cycle + 1, leaving cycle with the original value.
Part 2
Looking at the main. Let's reformat it a little bit.
int main()
{
unsigned long int i=0,j=0,loop=0;
Change these to long. By the way %ld in scanf expects long anyway, not unsigned long.
while (scanf("%ld %ld",&i,&j) != EOF)
Be careful with scanf: it has more return values than just EOF. Scanf will return EOF if it is not able to make a conversion. If it is able to scan one number, but not the second one, it will return 1. Basically a better test here is != 2. If scanf does not return two, something went wrong with the input.
{
if(i > j)
{
unsigned long int t=i;i=j;j=t;
}
for(loop=i;loop<=j;loop++)
{
cycle=0;
cycle=calculate(loop );
if(cycle>final)
{
final=cycle;
}
}
calculate is called hailstone_sequence_length now, and so this block can just have a local variable: { long len = hailstone_sequence_length(loop); if (len > final) final = len; }
Maybe final should be called max_length?
printf("%ld %ld %ld\n",i,j,final);
final=0;
final should be a local variable in this loop since it is separately used for each test case. Then you don't have to remember to set it to 0.
}
return 0;
}

Resources