guys... I am kinda new in programming, so I would ask you this : I ´ve made program for user that would like to write down position of something. I assume , that this should be done using array. It looks like this :
#include <stdio.h>
int main()
{
int positionA[1][1];
int i;
int j;
printf("Insert postitions x and y : \n");
for ( i=0;i<2;i--)
{scanf("%d", &i);
for ( j=0;j<2;j--)
{
scanf("%d", &j);
}
printf("%d", positionA[i][j]);
}
return 0;
}
Please, dont be mad on me ,if I have this program made bad.
Well and problem is, that I insert numbers like 2,5 . And at the output it appears like 1695956464 .
Sorry for bad English :P. And thanks for help !
int positionA[1][1];
Declares a 1-by-1 array. That is, positionA[0][0] is its only element. You later address positionA[1][1], which is outside its bounds. In C, an array with a size of, say, 10, allows indexes from 0 to 9 (that's 10 elements, starting with zero).
Also, NEVER mess with your loop variables inside the loop:
scanf("%d", &j);
assigns the input value to the variable j, which is your loop index, causing you to loop unknown number of times, and access the array way out of bounds.
Related
I have made a dynamic array of integers in C, here is my code
#include <stdio.h>
int main(){
int count=0, i, input;
int *myarr;
myarr=(int*)malloc(4*sizeof(int));
while(1){
scanf("%d", &input);
myarr[count]=input;
count++;
if (input == -1) break;
}
for (i=0; i<count; i++){
printf("%d ", myarr[i]);
}
return 0;
}
From the code, I thought i clearly made an array of 4 integers only i.e myarr[0] up to myarr[3], how come when i insert even 10 integers, it still prints all of them, it doesn't print garbage as i thought it would after the fourth integer... Maybe i didn't understand the point of dynamic creating an array?? Make me straight please!
You should only access myarr[0] up to and including myarr[3].
Accessing any other index is undefined behaviour: it might work, it might not.
Also, myarr[count]==input looks like a typo. Did you mean myarr[count] = input? The way you have it is testing if myarr[count] equals input. Technically the way you have it is undefined behaviour for any element of myarr since you are making use of uninitialised data.
I want to make a spiral in C using a 2D matrix, such as the one shown below:
This is the code that I worked out. But it keeps going into an infinite loop. I can't seem to get an output. Can anyone tell me what mistake I'm making in this logic?
And I know its pretty cumbersome, but the assignment is to get the output for any "n" dimensional array, and we need to use all the row_left,row_right, etc variables, as they're given in the question.
#include<stdio.h>
int main(void)
{
int array[6][6]={1},dim,row_right=0,row_left=dim-1,col_up=0,col_down=dim-1;
int i,j,num,cnt;
printf("Enter the dimensions of 2D array:\n");
scanf("%d",&dim);
num=dim*dim;
cnt=0;
while(cnt!=num)
{
for(j=col_up;j<=col_down;j++)
{
if(j=0)
array[row_right][j]=1;
else
array[row_right][j]=array[row_right][j-1]+1;
}
for(i=row_right+1;i<=row_left;i++)
array[i][col_down]=array[i-1][col_down]+1;
for(j=col_down-1;j>=col_up;j--)
array[row_left][j]=array[row_left][j+1]+1;
for(i=row_left-1;i>row_right;i--)
array[i][col_up]=array[i+1][col_up]+1;
row_right++;
row_left--;
col_up++;
col_down--;
cnt++;
}
for(i=0;i<dim;i++)
{
for(j=0;j<dim;j++)
printf("%d\t",array[i][j]);
printf("\n");
}
return 0;
}
if(j=0)
is almost surely wrong. This sets j to zero and always evaluates to a false condition. The correct condition uses j == 0.
Also, the code uses the variable dim before it is read by scanf.
You forgot to initialize the variable dim. That is used in following line :
int array[6][6]={1},dim,row_right=0,row_left=dim-1,col_up=0,col_down=dim-1;
With correctly formatted code you probably would have seen this.
//Program Written By: Andre Chitsaz-zadeh
//Program Written On: 10/7/12
//Program calculates book cost for multiple book orders. Program written using multiple functions.`
#include <stdio.h>
#define SIZE 5
void inputData();
int main ()
{
inputData();
}
void inputData()
{
int i = 0;
int costs[5];
printf( "\nPlease enter five products costs.\n" );
while(i < 5)
{
scanf("%d", costs[i]);
i = i + 1;
}
}
Why do I get in exception error? The program looks simple enough! It compiles without problems but as soon as I input a number it says that "this program has stopped working". Thanks!!
scanf("%d", costs[i]);
Should be:
scanf("%d", &costs[i]);// &cost[i] gets the address of the memory location you want to fill.
It should be
while(i < 5) {
scanf("%d", &costs[i]);
i = i + 1;
}
a little typo I assume, anyway you need to provide the address of the element of the array you want to scan the integer into.
I would guess it's this line:
scanf("%d", costs[i]);
It should be:
scanf("%d", &costs[i]);
scanf needs a pointer to the variable in which it should put the read result.
This looks like a homework question, judging by the comment about the program have multiple functions. If functions are new, then pointers have probably not been covered yet. In that case, read my explanation as:
scanf needs a & before the variable in which it should put the read result. You'll learn why in a few weeks.
I think you'll need to change you scanf line to
scanf("%d", &costs[i]);
You need to pass the address of the int to have user input written to it. Your current code passes the value of costs[i]. This is undefined so will point to an unpredictable, and probably not writeable, location in memory.
I'm just a beginner at C.
I'm trying to make a simple program to arrange the user-entered digits in ascending order. I have figured out the solution but can't understand why my other code wouldn't work :(
-------------------------------------------------------------------------
working code:
-------------------------------------------------------------------------
#include <stdio.h>
int main()
{
int i,j,num[10];
printf("Enter 10 numbers\n");
for (i=0;i<10;i++)
{scanf("%d",&num[i]);}
for (i=0;i<9;i++)
{
for (j=i+1;j<10;j++)
{
if (num[i]>num[j])
{
num[i]+=num[j];
num[j]=num[i]-num[j];
num[i]=num[i]-num[j];
}
}
}
printf("The numbers in ascending order are:");
for (i=0;i<10;i++)
{
printf(" %d",num[i]);
}
return 0;
}
-------------------------------------------------------------------------
code that won't work:
-------------------------------------------------------------------------
#include <stdio.h>
int main()
{
int i,j,num[10];
printf("Enter 10 numbers\n");
for (i=1;i<=10;i++)
{scanf("%d",&num[i]);}
for (i=1;i<10;i++)
{
for (j=i+1;j<=10;j++)
{
if (num[i]>num[j])
{
num[i]+=num[j];
num[j]=num[i]-num[j];
num[i]=num[i]-num[j];
}
}
}
printf("The numbers in ascending order are:");
for (i=1;i<=10;i++)
{
printf(" %d",num[i]);
}
return 0;
}
In the latter program, numbers appear out of order, and there even are numbers that haven't been entered.
My question is, isn't it basically the same code? Just that in the latter program numbers would be stored from num[1] to num[10] instead of num[0] through num[9]?
Does it have something to do with array definitions?
It seems I have serious misconceptions, please help me out!
In C, when you have int num[10];, your indexes need to go from 0 to 9, never to 10. So look over your code, if any i or j ends up with a value of 10 any time during the program run, that's bad news.
Indexes in C go start from 0. so when you declare an array of size 10, and you try to get element at index 10, you're actually getting the 11th element. Since you haven't defined the 11th element, the array will most likely get some random numbers from memory, which is why you are noticing numbers you have note entered.
Since you are new to programming, I would suggest taking the time now to really learn about how C manages memory, and how different data structures access the memory. It might be a little boring now, but you'll save yourself some headaches in the future, and you will start to build good habits and good practices, which will lead to writing good, optimal code
for(i=0;i<9;i++) //**i<9**
for (j=i+1 ...)
If i=8 then j=9 , everything is OK.
In second code snippet:
for(i=0;i<10;i++) //**i<10**
for (j=i+1 ...)
If i=9 then j=10, so you try to access num[10] and it gives you error.
If you want to access num[10] then you must declare array int num[11] and then you can access num[10].
Array Basics
int num[10]
Capacity of array = 10
Every element of this array are integer.
First element index is 0. So If you want to access first element , num[0]
Last element index is 9. So If you want to access last element, index must be length of array - 1 , so num[9]
There are 10 elements in the array and they are :
num[0] , num[1] , num[2] , num[3] , num[4] , num[5] , num[6] , num[7] , num[8] and num[9]
You can learn further at http://www.cplusplus.com/doc/tutorial/arrays/
In your non-working example, you have invalid memory accesses. For example, when i = 9 and j = 10, you access num[10], which is invalid memory.
Welcome to programming! I believe you are a bit confused about how arrays are indexed in C.
Arrays in most languages (including C) are known as zero-indexed arrays. This means the start of an array is always at position 0. So if you make int num[10], trying to access num[10] isn't actually a valid call at all, because the start is num[0]. Hence you can only access from num[0] to num[9].
It's an easy mistake to make, even though I've been programming for years sometimes when it's been a long night I'll still make silly array indexing issues.
In your other code example, you were doing this:
int num[10];
for(int i = 1; i <= 10; i++) {
//...
}
That means you have an array with ten spaces [0-9], and for the last part of your for loop, you were trying to access num[10] which doesn't exist.
Your working example goes from 0-9 and never tries to read num[10], so it works.
Arrays in C, as in most languages, start with position 0 and count that as position one. So the last element of your array would be the size you entered when you declared the variable, minus one.
My teacher gave an assignment to me. The question is below:=
Write a program that prompts the user to enter 10 double numbers. The program should accomplish the follwing:
a. Store the information in a 10-element array.
b. Display the 10 numbers back to the user.
I could do all of the above in main().
Hint: You should use loops, not hardcode the values 0 through 9. It should be easy to convert your program to accept 1000 numbers instead of 10.
For a bonus mark, do at least one of the tasks (a or b) in a separate function. Pass the array to the function; do NOT use global (extern) variables.
I confused above. I wrote a program in the source code. Am I doing wrong? It is below:=
#include<stdio.h>
int main(void)
{
int number[10];
int i;
for (i = 0; i <10; i++)
printf("%d.\n", i, number[i]);
printf("\n\nPress [Enter] to exit program.\n");
fflush(stdin);
getchar();
return 0;
}
Thanks.
Not too bad so far, I'd like to make the following comments:
if you need to input double numbers, you should probably use double rather than int.
you need a statement (maybe in your current loop but possibly in another loop preceding the current one) which inputs the numbers. Look into scanf for this.
Using %d with printf is for integers, not doubles. You will have hopefully already figured out the format string to used when you looked into scanf above.
Bravo for using the correct int main(void) form and for not including conio.h :-)
Once you've figured those bits out, then you can worry about doing it in a separate function.
Based on the code you have given above, I would suggest reading up on the following:
scanf
functions in C, particularly passing arrays to functions: this link should be good.
Note to OP: If you were able to do (a) and (b) in main(), the code above is not complete. It would be nice the functions you created for getting (a) and (b) above done for getting to the root of your "confusion".
Let me know in case you need more help.
HTH,
Sriram
Try this it may sloves your problem.
#include<stdio.h>
int main(void)
{
double number[10];
int i;
printf("Enter double numbers:");
for (i = 0; i <10; i++)
scanf("%lf",&number[i] );
printf("The numbers you entered are:");
for (i = 0; i <10; i++)
printf("%lf\n",number[i] );
return 0;
}