I've just started to learn coding
#include <stdio.h>
int main(void)
{
int i=0, x=0;
for (i=1; i<=100; i++) {
x++;
if (x%5==0 || x%3==0)
printf("The numbers are : %d\n", &x);
}
return 0;
}
so I'm trying to print all the integers <=100 that are divisible by either 3 or 5.
In the line:
printf("the numbers are : %d", &x);
printf is expecting an integer because of the %d format specifier, you have given it the address of an integer, thats what the & means, address of x. To fix this give printf what it desires, an int:
printf("the numbers are : %d", x);
The ampersand (&) operator returns the address of a variable, which isn't what you want - you just want the value. Also, note that you don't need two variables (x and i) - you can just use the loop's counter:
printf("The numbers are:\n");
for (i = 1; i <= 100; i++) {
if (i % 5 == 0 || i % 3 == 0) {
printf("%d ", i);
}
}
Removing the & will eliminate the error. You may also want to move the "the numbers are:" statement outisde of the loop so the text does not print each time through the loop.
Related
#include <stdio.h>
#include <stdlib.h>
int main() {
int i;
int mult;
int n;
int ans;
ans = mult * i;
printf("Please enter a multiple you want to explore.");
scanf("%d", &mult);
printf("Please enter the number which you would want to multiply this number till.");
scanf("%d", &n);
for(i = 0; i<n; i++) {
printf("%d x %d = %d \n", &mult, &i , &ans);
}
return 0;
}
Hi guys, this is a simple code which is supposed to help the user to list the times table for n times. However, i am receiving undefined behaviour and I am quite stumped as to what is wrong with my implementation of my "for" loop.
I am receiving this as my output.
6356744 x 6356748 = 6356736
for n times in my consoles.
I want to ask
Is anything wrong with the logic of my code? (i assume i do have a problem with my code so please do enlighten me)
Would it be better(or even possible) to use pointers to point to the memory addresses of the mentioned variables when i have to change the value of the variables constantly? If yes, how do i go around doing it?
Thanks!
In printf you must provide integers. You are now giving the addresses of integers. So change
printf("%d x %d = %d \n", &mult, &i , &ans);
to
printf("%d x %d = %d \n", mult, i, ans);
and to make the table, replace ans with just mult*i, so:
printf("%d x %d = %d \n", mult, i, mult*i);
You should also check the return value of scanf to check if it has succeeded reading your input:
do {
printf("Please enter a multiple you want to explore.");
} while (scanf("%d", &mult)!=1);
do {
printf("Please enter the number which you would want to multiply this number till.");
} while (scanf("%d", &n)!=1);
The things you see are the values of the variables memory location.
Change your lines inside for loop as below
ans = mult * i;
printf("%d x %d = %d \n", mult, i, ans);
There are some mistakes in your code .
you are using the & operator in print statement which is used to print the address of the variable.
Initiate the loop with the value '1' instead of '0' & execute the loop till 'i' less than equal to 'n'.
instead of using the ans variable outside the loop , use it inside the loop as it evaluate the multiplication result in each iteration of the loop.
#include <stdio.h>
int main()
{
int i;
int mult;
int n;
int ans;
printf("Please enter a multiple you want to explore.");
scanf("%d", &mult);
printf("Please enter the number which you would want to multiply this number till.");
scanf("%d", &n);
for(i = 1; i<=n; i++) {
ans = mult*i ;
printf("%d x %d = %d \n", mult, i , ans);
}
return 0;
}
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Closed 6 years ago.
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I'm not sure what i'm doing wrong but the for loop is not initializing
The code just goes immediately to displaying the printfs. That have no values in them since the for loop didn't activate
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#define PAUSE system("Pause")
main() {
// INITALIZE VARIABLES
int number = 0;
int i = 0;
int odd = 0;
int even = 0;
int totalNum = 0;
int tempNum = 0;
int count;
printf("Enter a number between 2 and 25\n");
scanf("%i", &number);
do{
if (number < 2 || number > 25)
printf("That was an invalid number please try again\n");
scanf("%i", &number);
} while (number < 2 || number > 25);
printf("Enter how many numbers you want to input\n");
scanf("%i", &count);
for (i = 1; i == count; ++i){
printf("input numbers\n");
scanf("%i", &tempNum);
if (tempNum % 2 == 0)
even++;
else
odd++;
totalNum = totalNum + tempNum;
} // END FOR LOOP
// DISPLAY OUTPUT
printf("You entered %i numbers\n", count);
printf("The sum of the %i numbers is %i\n", count, totalNum);
printf("The average of the %i numbers is %i\n", count, totalNum / count);
printf("You entered %i odd numbers and %i even numbers\n", odd, even);
PAUSE;
} // END MAIN
Your loop will only execute at best once, when count == 1 as you initialize i to 1.
If you enter a 1 for count,
printf("Enter how many numbers you want to input\n");
scanf("%i", &count);
the loop will run exactly once, until i increments to 2
You probably want:
for (i = 1; i <= count; ++i){
do{
if (number < 2 || number > 25)
printf("That was an invalid number please try again\n");
scanf("%i", &number);
} while (number < 2 || number > 25);
it should be...
do{
if (number < 2 || number > 25){
printf("That was an invalid number please try again\n");
scanf("%i", &number);
}
} while (number < 2 || number > 25);
else it asks always another number
i = 1, so i == count; gives false therefore the loop is ignored.
A for loop in C works like this:
for ( variable initialization; condition; variable update ) {
/* Do something... */
}
The loop will execute for as long as condition is true. So, when you do:
for (i = 1; i == count; ++i)
The loop will execute for as long as i == count is true. So, unless count holds 1 when this line is executed, the loop will never run.
As others pointed out, you probably want this:
for (i = 1; i <= count; ++i)
So your loop will run for all values of i, until it reaches count.
As a side note i should point out that the usual way to write for loops in C is something like this:
for (i = 0; i < count; i++)
We start with i = 0 because C arrays are zero-based, so the Nth element of an array has index n-1
You were so close. In addition to fixing your loop test clause for (i = 1; i <= count; i++), I would suggest using " %d" for your format specifier. Your do loop need only be a while loop to avoid printing your invalid number message every time.
Additionally, While not an error, the standard coding style for C avoids caMelCase variables in favor of all lower-case. See e.g. NASA - C Style Guide, 1994.
With those changes, (and changing your odd/even check to a simple &) you could write your code as follows.
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
// #define PAUSE system("Pause")
int main (void)
{
int number, i, odd, even, totalnum, tempnum, count;
number = i = odd = even = totalnum = tempnum = count = 0;
printf ("enter a number between 2 and 25: ");
scanf (" %d", &number);
while (number < 2 || number > 25) {
printf ("invalid number, again (between 2 and 25): ");
scanf (" %d", &number);
}
printf ("numbers to input: ");
scanf (" %d", &count);
for (i = 1; i <= count; i++) {
printf ("input number %2d: ", i);
scanf (" %d", &tempnum);
if ((tempnum & 1) == 0)
even++;
else
odd++;
totalnum = totalnum + tempnum;
}
printf ("You entered %d numbers\n", count);
printf ("The sum of the %d numbers is %d\n",
count, totalnum);
printf ("The average of the %d numbers is %d\n",
count, totalnum / count);
printf ("You entered %d odd numbers and %d even numbers\n",
odd, even);
// PAUSE;
return 0;
}
note: main is type int (e.g. int main (int argc, char **argv) or simply int main (void) to indicate no arguments taken). Since it is type int it will return a value to the shell. While historic implementations may have allowed void main that is no longer the case for portable code.
Example Use/Output
$ /bin/forskipped
enter a number between 2 and 25: 4
numbers to input: 4
input number 1: 1
input number 2: 2
input number 3: 3
input number 4: 4
You entered 4 numbers
The sum of the 4 numbers is 10
The average of the 4 numbers is 2
You entered 2 odd numbers and 2 even numbers
Look it over and let me know if you have any questions.
My console keeps on crashing after entering a few numbers. I am trying to get an array of 10 numbers from the user thru the console and then taking count of positives, negatives, evens, and odds. What am I doing wrong?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
int pos, neg, even, odd;
int nums[10];
printf("Give me 10 numbers: ");
pos = neg = even = odd = 0;
for(int i = 0; i < 10; i++){
scanf(" %d", nums[i]);
if(nums[i] > 0){
pos++;
if(nums[i] % 2 == 0){
even++;
}
else{
odd++;
}
}
else{
neg++;
}
}
printf("Positives: %d, Negatives: %d, Evens: %d, Odds: %d\n", pos, neg, even, odd);
return 0;
}
In your code,
scanf(" %d", nums[i]);
should be
scanf(" %d", &(nums[i]));
or,
scanf(" %d", nums+i);
as you need to pass the pointer to variable as the format specifier's argument in scanf() .
To elaborate, %d expects a pointer to int and what you're supplying is an int variable. it invokes undefined behavior.
That said,
Always check the return value of scanf() to ensure proper scanning.
int main() should be int main(void) to conform to the standard.
Modify scanf like scanf(" %d", &nums[i]);
scanf(" %d", nums[i]);
Scanf expects a pointer to a location to write to, and you're not giving it one.
Change your scanf to:
scanf(" %d", &(nums[i]));
to make your program work.
With this change I tested your program with stdin of
20 10 9 1 39 1 2 2 31 1
And recieved output:
Give me 10 numbers: Positives: 10, Negatives: 0, Evens: 4, Odds: 6
ideone of the thing for your testing purposes.
Change scanf(" %d", nums[i]); to scanf(" %d", &nums[i]);, because scanf() needs addresses. The parentheses around nums[i] isn't necessary, and may effect readability.
Also note that 0 is even, but not negative.
When scanf is usedto convert numbers, it expects a pointer to the corresponding type as argument, in your case int *:
scanf(" %d", &nums[i]);
This should get rid of your crash. scanf has a return value, namely the number of conversions made or the special value EOF to indicate the end of input. Please check it, otherwise you can't be sure that you have read a valid number.
When you look at your code, you'll notice that you don't need an array. Afterreading the number, you don't do aything with the array. You just keep a tally of odd, even and so on numbers. That means you just need a single integer to store the current number. That also extends your program nicely to inputs of any length.
Here's a variant that reads numbers until the end of input is reached (by pressing Ctrl-D or Ctrl-Z) or until a non-number is entered, e.g. "stop":
#include <stdlib.h>
#include <stdio.h>
int main()
{
int count = 0;
int pos = 0;
int neg = 0;
int even = 0;
int odd = 0;
int num;
while (scanf("%d", &num) == 1) {
count++;
if (num > 0) pos++;
if (num < 0) neg++;
if (num % 2 == 0) even++;
if (num % 2 != 0) odd++;
}
printf("%d numbers, of which:\n", count);
printf(" %d positive\n", pos);
printf(" %d negative\n", neg);
printf(" %d even\n", even);
printf(" %d odd\n", odd);
return 0;
}
Change scanf statement after for loop to
scanf(" %d", &nums[i]);
I have a problem, I tried to write a program to show the whole sum from 1 to 22 and after that, to do 2 while loops. The first one is supposed to perform the sum of some numbers given by the user, as an example: you type 10, 30 and 40 then as you enter a 0 the program sums the first three numbers. Unfortunetly the first while loop is not working. It goes directly to the last while loop where it is supposed to type a decimal numbers like (10.20 30.50 40.55) and after you type 0 again it sum those numbers and add and multipli every entry with 1.19. So far the last loop is working properly, unfortunately the second loop does not, if I move printf and scanf over the while it let me write but just start writing w/o stopping the number I wrote . Thank You in advance!
Here is the code :
#include <stdio.h>
int main()
{
int sum = 0;
int a;
int b;
double i;
double sum1 = 0;
for (a= 0; a <= 22; a++) {
sum = sum + a;
printf("the sum from 1 till 22 : %i\n ", sum);
}
while (b != 0) {
printf("type a number:");
scanf("%i", &b);
sum += b;
printf("%i\n", b);
}
printf("the sum is : %i\n", sum);
while(i !=0) {
printf ("Type a decimal number:");
scanf ("%lf",&i);
sum1 += i*1.19;
printf("%lf\n", i);
}
printf("The decimal summ is: %lf\n",sum1);
return 0;
}
You don't initialise i to any value before entering the loop with
while(i != 0)
i might very well be zero at this point, so your loop won't be entered even once. Initialising i to a non-zero value should fix this particular problem. The same holds for the variable b.
You should turn on warnings in your compiler, so it can show you problems like this one.
The first time the condition of the second while is evaluated, b has undefined value, since it wasn't initialized. The same applies to the third while.
Whether or not both loops are executed is only a question of chance.
Initialize both variables with non-zero values to ensure both whiles are entering. Or use a do-while:
do {
printf("type a number:");
scanf("%i", &b);
sum += b;
printf("%i\n", b);
} while (b != 0);
Don't test b with while, test it after the user enters the number. Then you can use break to exit the loop.
while (1) {
printf("type a number:");
scanf("%i", &b);
if (b == 0) {
break;
}
sum += b;
printf("%i\n", b);
}
while(1) {
printf ("Type a decimal number:");
scanf ("%lf",&i);
if (i == 0.0) {
break;
}
sum1 += i*1.19;
printf("%lf\n", i);
}
Your only issues are initialization: see edits in the code below. (it compiles and runs)
Did you get any compiler warnings for these? If not, you should change your settings so you do.
#include <stdio.h>
int main()
{
int sum = 0;
int a;
int b=-1; //initialize (any non-zero value will work)
double i;
double sum1 = 0;
for (a= 0; a <= 22; a++) {//a initialized in for(...) statement, (this is good)
sum = sum + a;
printf("the sum from 1 till 22 : %i\n ", sum);
}
while (b != 0) { //b Needs to be initialized before using (done above)
printf("type a number:");
scanf("%i", &b);
sum += b;
printf("%i\n", b);
}
printf("the sum is : %i\n", sum);
i=-1; //initialize i to any non-zero value
while(i !=0) {
printf ("Type a decimal number:");
scanf ("%lf",&i);
sum1 += i*1.19;
printf("%lf\n", i);
}
printf("The decimal summ is: %lf\n",sum1);
getchar();
return 0;
}
Is there a way to select multiple elements in array using one line of code in C? For instance, say I had the following code (assuming I already asked the user for twenty numbers, the first ten I asked to be positive and the last ten I asked to be negative):
if (myArray[0 through 9] > 0)
{
printf("Thank you for providing positive numbers!");
}
else
{
printf("Sorry, please try again!");
}
if (myArray[10 through 19] < 0)
{
printf("Thank you for providing negative numbers!");
}
else
{
printf("Sorry, please try again!");
}
What code could I substitute for "through"? I am fairly new to this language, and have never heard of a way of doing so. I know that with this particular code I could make two arrays, one for the positive numbers and one for the negative numbers, but I am curious to know for other programming projects.
Thank you for reading and answering!
There's nothing built-in that does it, you need to write a loop. Don't forget that array indexes start at 0.
int all_positive = 1;
int i;
for (i = 0; i < 10; i++) {
if (myArray[i] <= 0) {
all_positive = 0;
break;
}
}
if (all_positive) {
printf("Thank you for providing positive numbers!\n");
}
int a[20];
// entering values for the array
_Bool first_positive = 1;
for ( size_t i = 0; i < 10 && first_positive; i++ )
{
first_positive = 0 < a[i];
}
if ( first_positive ) puts( "Hura, first 10 elements are positive" );
_Bool last_negative = 1;
for ( size_t i = 10; i < 20 && last_negative; i++ )
{
last_negative = a[i] < 0;
}
if ( last_negative ) puts( "Hura, last 10 elements are negative" );
Instead of type name _Bool you can use type int if your compiler does not support _Bool
The program requests number of rows (1D array).
Then asks for 2 integers, whole numbers.
Then asks user to select 2 rows.
The sum of the 2 selected rows is then added.
#include <stdio.h>
int main ()
//1D_Array. Load Element and Add Sumline .
//KHO2016.no5. mingw (TDM-GCC-32) . c-ansi .
{
//declare
int a,b,c,d,e,sum1=0;
int array[50];
int i,j,elm1,elm2;
//valuate
printf ("Plot a number of elements [1 - 20]: ");
scanf ("%d",&a);
printf ("Plot a value : ");
scanf ("%d",&b);
printf ("Plot an increment value : ");
scanf ("%d",&c);
//calculate
{for (i<0;i<=a;i++)
{array[i] =(b+(++c)); // set value for variable [i], inside the array subscript. the vairable [i] must have an INT, and an increment to function !
sum1 = (sum1 + array[i]);
printf ("Row [%.2d] : %.2d + %.2d = %d\n",i,b,c,array[i]);}
printf ("\nSum total = %d\n",sum1);}
printf ("\nRow [%.2d] = %d\n",b,array[b]);
printf ("Row [%.2d] = %d\n",a,array[a]);
printf ("Select 2 Rows :\n");
scanf ("%d%d",&elm1,&elm2);
d=elm1;
e=elm2;
printf ("You selected Row [%.2d] = %d\n",d,array[d]);
printf ("You selected Row [%.2d] = %d\n",e,array[e]);
printf ("The sum of two selected Rows [%d]+[%d] : %d + %d = %d\n",d,e,array[d],array[e],array[d]+array[e]);
//terminate
return 0;
}