What's the differences between the next three codes.
1:
struct node
{
int data;
}tree[100050];
/*
some codes
*/
int main()
{
int tot;
scanf("%d",&tot);
int tmp;
for(int i=0;i<=tot;++i){
scanf("%d",&tmp);
tree[i].data=tmp;
insert(i,1);
}
}
Wrong Answer
2:
struct node
{
int data;
}tree[100050];
/*
some codes
*/
int main(){
int n;
scanf("%d",&n);
int tmp;
for(int i=0;i<=n;++i){
scanf("%d",&tree[i].data);
insert(i,1);
}
}
Accepted
3:
struct node
{
int data;
}tree[100050];
/*
some codes
*/
int main()
{
int tot;
scanf("%d",&tot);
int tmp;
for(int i=0;i<=tot;++i){
scanf("%d",&tmp);
tree[i].data=tmp;
insert(i,1);
tmp=0;
}
}
Accepted
The first codes can not pass all tests, but next two codes can pass all tests.
The problem is here POJS024。It is written in Chinese.
Input is a list of number which builds a Binary Sort Tree，the first number is the root.
Output the In-Order Traversal and the Post-Order Traversal。
All three use wrong logic. You got lucky in the second and third, probably because insert has some special handling of values that are equal to 0.
The mistake in all three is that you are using
for(int i=0;i<=n;++i){
instead of
for(int i=0;i<n;++i){
^^^ Needs to be < not <=
If you had code to check the return value of scanf, you would have caught your error sooner.
if ( scanf(...) == 1 )
{
insert(i,1);
}
That's why it's very important that you make a habit of checking the return value of scanf family of functions ALWAYS to make sure that you were able to read all data that you were expecting to.
The problem (which actually exists in some sense in all three cases) is that you call scanf() one too many times because you're using a <= comparison in the for loop instead of the using the < comparison. As a result the last call to scanf() actually fails. and you call insert(i,1) one more time than it should be called.
In example 1 that results in tree[i].data being set to the last value that was read into tmp, in examples 2 and 3 that value is left at zero. Apparently the extra call to insert(i,1) doesn't cause a problem in that particular scenario.
Related
This is a problem where you have to enter a number having non zero digits and the program will convert it into a IP address having 4 parts and each part less than 255, and have to print all the IP addresses.
I have tried this recursive method and am running into an infinite loop.
#include<stdio.h>
#include<math.h>
int a[3];
void comuni(unsigned long n,int count){
int i=count;
do{
if(count<0){
return;
}
int t=pow(10,i);
a[count]=n/t;
int rem=n%t;
if(a[count]<=255 &&a[count]>0 && count>=0){
printf("%d.",a[count]);
comuni(rem,count-1);
}
i++;
}while(1);
}
int main()
{
// Insert your code here.
unsigned long n;
scanf("%ul",&n);
comuni(n,3);
return 0;
}
Your question is why there is an infinite loop.
Starting with a look at while(1) it does not surprise me.
But wait, there is a return inside, which is used if count is less than 0.
Nothing inside the loop changes count. So if count is greater or equal 0 to begin with you got your infinite loop.
I think the problem which causes that situation is that this
int i=count;
does NOT cause count to change when you change i.
You do change i inside the loop, but you know - it does not help.
I have done my fair share of studying the C language and came across this inconsistency for which I cannot account. I have searched everywhere and reviewed all data type definition and relational syntax, but it is beyond me.
From the book C How to Program, there is a question to make a binary to decimal converter where the input must be 5-digits. I developed the follow code to take in a number and, through division and remainder operations, split it into individual digits and assign each to an element in of an array. The trouble arises when I try to verify that the number entered was indeed binary by checking each array element to see whether it is a 1 or 0.
Here is the code:
#include <stdio.h>
int power (int x, int y); //prototype
int main(void)
{
int temp, bin[5], test;
int n=4, num=0;
//get input
printf("%s","Enter a 5-digit binary number: ");
scanf("%d", &temp);
//initialize array
while(n>=0){
bin[n]=temp/power(10,n);
temp %= power(10,n);
n--; }
//verify binary input
for (test=4; test>=0; test--){
if ((bin[n]!=0)&&(bin[n]!=1)){
printf("Error. Number entered is not binary.\n");
return 0; }
//convert to decimal
while(n<=4){
num+=bin[n]*power(2,n);
n++; }
printf("\n%s%d\n","The decimal equivalent of the number you entered is ",num);
return 0;
}
//function definition
int power(int x, int y)
{
int n, temp=x;
if(y==0) return 1;
for(n=1; n<y; n++){
temp*=x; }
return temp;
}
Could someone explain to me why regardless of input (whether: 00000, or 12345), I always get the error message? Everything else seems to work fine.
Thank you for your help.
Update: If the if statement is moved to the while loop before. This should still work right?
Update2: Never mind, I noticed my mistake. Moving the if statement to the while repetition before does work given the solution supplied by sps and Kunal Tyagi.
After this
while(n>=0){
bin[n]=temp/power(10,n);
temp %= power(10,n);
n--; }
n is set as -1 so when you try to convert to decimal the statement bin[n] is actually bin[-1] so it returns you error.
One issue is that, while checking if the number is binary or not, you are returning at wrong place. You need to return only if the number is not binary. But you are returning outside the if condition. So your program returns no matter what the input is.
for (test=4; test>=0; test--){
if ((bin[test]!=0)&&(bin[test]!=1))
printf("Error, numbered entered was not binary.\n");
// Issue here, you are returning outside if
return 0; } //exit program
You can change that to:
for (test=4; test>=0; test--){
if ((bin[test]!=0)&&(bin[test]!=1)) {
printf("Error, numbered entered was not binary.\n");
// Return inside the if
return 0; // exit program
}
}
There is one more issue. Before you convert your number to decimal, you need to set n = 0;
//convert to decimal
n = 0; /* need to set n to zero, because by now
n will be -1.
And initially when n is -1, accessing
bin[-1] will result in undefined behavior
*/
while(n<=4){
num+=bin[n]*power(2,n);
n++; }
This looks like a homework, but your issue is in the brackets. More specifically line 23. That line is not part of the logical if statement despite the indentation (since that doesn't matter in C).
No matter what, the program will exit on test=4 after checking the condition.
Solution:
if ((bin[test]!=0)&&(bin[test]!=1)) { // << this brace
printf("Error, number entered was not binary.\n");
return 0; } } //exit program // notice 2 braces here
I am trying to write a simple program. I am a begineer and i am not getting a value to total. When i am trying to print . I am getting a address as output . Can anyone explain me what is the mistake and correct my program .
#include<stdio.h>
void main()
{
int first,second,total;
printf("enter the value for the first");
scanf("%d",&first);
printf("enter the value for the second");
scanf("%d",&second);
total=power(first,second);
printf("The value for power is %d",power);
}
int power(int doom1,int doom2)
{
int temp=doom1;
int i;
for(i=1;i<=doom2;i++)
{
temp=temp*doom1;
}
return temp;
}
You are printing the wrong variable:
total=power(first,second); //here you are getting return value in variable total
printf("The value for power is %d",power); // power is the function name not variable
Replace this line with:
printf("The value for power is %d",total); // you need to print `total`
Also you have to declare your function prototype before main():
int power(int ,int);
and you should use int main():
int main()
{
// your code
return 0;
}
In addition to passing total to printf instead of power, as you are just starting, make a point to always give your variables an initial value (initialize them). This prevents an attempt to read from uninitialized space which is the bane of new C programmers. (it will save you a lot of headaches). Attempting to read from an uninitialized variable is Undefined Behavior. That can result in anything from slipping by unnoticed, to causing your program to crash. It is to be avoided.
Also, as I explained in the comment, in C, the function main() is type int and it returns a value to its caller (usually the shell, or another program). When using main without arguments, the proper form is:
int main (void)
When accepting arguments, the proper form is:
int main (int argc, char **argv)
In either case, it should return a positive value upon completion. A return 0; at the end is all that is required. exit (0); is another function you can use to return a value. You will also see the form of main with arguments written as:
int main (int argc, char *argv[])
The first and second forms are the practical equivalents of each other, the first recognizing that an array passed to a function in C will decay to a pointer. But for now, just understand that they are equivalent.
You also have an error in your my_power calculation. int temp = doom1; should be int temp = 1; Your calculation was returning a value twice the actual product.
Your style of syntax is up to you, but I would suggest that expanding your syntax a little by using discretionary spaces and lines will make your code much more readable and make finding errors a bit easier. Here is an example regarding all of these points:
#include <stdio.h>
int my_power (int doom1, int doom2);
int main (void)
{
int first = 0; /* Always initialize your variable to prevent */
int second = 0; /* an inadvertant read from an unitialized */
int total = 0; /* value which is Undefined Behavior (bad). */
printf ("\n enter the value for the first : ");
scanf ("%d",&first);
printf (" enter the value for the second: ");
scanf ("%d",&second);
total = my_power (first,second);
printf ("\n The value for my_power is: %d\n\n", total);
return 0;
}
int my_power (int doom1, int doom2)
{
int temp = 1;
int i = 0;
for (i = 1; i <= doom2; i++)
temp = doom1 * temp;
return temp;
}
Output
$ ./bin/simple_function
enter the value for the first : 2
enter the value for the second: 7
The value for my_power is: 128
you are trying to print "power" without parameter
printf("The value for power is %d",power);
you should do
printf("The value for power is %d",total);
or
printf("The value for power is %d",power(first,second));
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
int prime(long long int);
long long int *arr; //array to hold n prime numbers
int main()
{
int i,count=4;;
long long int n;
scanf("%lli",&n);
arr=malloc(sizeof(long long int)*n);
arr[0]=2;
arr[1]=3;
arr[2]=5;
arr[3]=7;
if (n==1) printf("%lli",arr[0]);
else{ if (n==2) printf("%lli",arr[1]);
else{ if (n==3) printf("%lli",arr[2]);
else{ if (n==4) printf("%lli",arr[3]);
else
{
for(i=2;count<n;i++)
{
if(prime(6*i-1)) { /*As prime nos are always 6k+1 or
arr[count]=6*i-1; 6k-1fork>=2 I checked only for those*/
count++; }
if(prime(6*i+1)&&count<=n) {
arr[count]=6*i+1;
count++; }
}
printf("%lli",arr[count]);
}}}}
//free(arr);
return 0;
}
int prime(long long int x)
{
int j=1,flag=1;
while(arr[j]<=sqrt(x))
{
if (x%arr[j]==0)
{
flag=0;
break;
}
j++;
}
return flag;
}
The code is working only for n=1,2,3,4, i.e i=0,1,2,3 for which the values are explicitly given. For n=5 onwards it is giving 0 as O/P
There is some glitch related to the global dynamic array as free(arr) is giving core dump error.
Q: Is this the right way to declare a global dynamic array? What could be the problem in this code?
Thank You in advance.
If that is your actual code you have 4 bugs:
2 line comment scopes out a line of your code
the second if should check count < n not count <= n as if count == n you cannot write to arr[count]
You cannot print arr[count] only arr[count-1] which is probably what you mean
In the case where n is less than 4 you still set arr[1], arr[2] and arr[3] which may be out of bounds
It is of course also inefficient to call sqrt(x) in every loop iteration, potentially you should call it outside and there may be a potential rounding issue bug due to the way square roots are calculated, so you might prefer:
while( arr[j] * arr[j] < x )
It would be preferable not to make this global and to pass it into your function.
It would also be preferable to move the main loop logic of your program outside of main().
I'm surprised you say you program works for n=1, 2 and 3 as it looks like you are setting out of bounds.
Your counter goes beyond the size of the array. Specifically both conditions (6i-1 and 6i+1) are met for i=2, and therefore counter is incremented twice, resulting in using arr[5] where you only allocated 5 places in the array. This is because you check counter<=n and not counter
Not sure this could be also be the reason for free creating a core dump, but it is possible (because once corrupting the memory, free may access corrupted data).
I'm writing a function for my homework which is supposed to tell if a given string is a palindrome or not.
Although I even tried it on paper with the word "otto", my program always returns 1.
Although this is a quite common question, I'd really like to know what I'm doing wrong instead of just copying a solution from here.
int is_palindrom(const char* palin)
{
int size = strlen(palin), i=0;
for (i=0;i<=(size/2); ++i)
{
if(palin[i] != palin[(size - i -1)])
{
return 1;
}
}
return 0;
}
Your code is correct, however please note that you may have an inverted logical expression. You are returning 1 in case of not equal, and 0 when it is. This means your function is working the opposite of "standard" C functions, where 1 evaluates to true.
Obviously, you are free to use whichever value you like to represent whatever you want. However, this can easily lead to confusion if someone else is reading your code. If bool is available, you should be using that; otherwise, you should always assume 1 is true and 0 is false.
Also, make sure to note is_palindrome takes a string and not an integer.
i.e. you must call it as is_palindrome("767") and not is_palindrome(767)
Your code does return 0 when it should. I am guessing when you read the string you pass as argument to your function, there are extra characters appended to the string, most probably a new line character. Try debugging the application or adding debug output in the function. For instance print the length of the string and the ascii codes of the characters in it.
Here is the code I used to verify it:
#include <stdio.h>
#include <string.h>
int is_palindrom(const char* palin)
{
int size = strlen(palin), i=0;
for (i=0;i<=(size/2); ++i)
{
if(palin[i] != palin[(size - i -1)])
{
return 1;
}
}
return 0;
}
int main(void) {
printf("%d", is_palindrom("otto"));
return 0;
}
Make sure your (const char *) has a "\0" at the end when you call this function.
#include<stdio.h>
#include<conio.h>
int is_palindrom(const char* jj);
int main(char *args){
int rr = is_palindrom("otto");
printf("rsult is %d", rr);
getch();
}
int is_palindrom(const char* palin)
{
int size = strlen(palin), i=0;
for (i=0;i<=(size/2); ++i)
{
if(palin[i] != palin[(size - i -1)])
{
return 1;
}
}
return 0;
}
I ran you code using above code snippet and it work fine for me.it returns 0 if palindrome is entered and 1 if entered value is not palindrome. the main part of the function is the loop
for (i=0;i<=(size/2); ++i) and the comparison if(palin[i] != palin[(size - i -1)]) the loop starts from 0 and then in condition palin[0] element and palin[4-0-1] i.e palin[3] element first o and last o in this case are mapped then the increement ++i takes place and then nest mapping of palin[second] and palin[second-last] elements happen so you can you either `++i' or 'i++'