Im trying to make a program which says how many times a specific digit appears on a 100 numbers sequence.
Meanwhile I got this error and I can´t understand what is the solution to this. I´d appreciate if you could get me some tip or the solution.
#include <stdio.h>
int main() {
int i, m, digit, val[99], count=0;
printf("Enter a number:");
scanf("%d", &val[0]);
while (val[0] < 0) {
printf("Enter a number:");
scanf("%d", &val[0]);
}
for (i=1;i<101;i++) {
val[i]=val[0]++;
printf("%d\n", val[i]);
}
printf("Enter a digit:");
scanf("%d", &m);
while (m<0||m>9) {
printf("Enter a digit:");
scanf("%d", &m);
}
do {
digit=val[i]%10;
val[i]=val[i]/10;
if (digit==m) {
count++;
}
}while (val[i]>0);
printf("The digit %d is printed %d times in this sequence.", m, count);
}
In the for loop you step outside of the array val of which the last index is 98. Instead of hard-coding the length of the array in several places it is more convenient to use a length macro, like this:
#define LEN(anArray) (sizeof (anArray) / sizeof (anArray)[0])
...
for (i = 1; i < LEN(val); i++) {
...
Also, in the do-while loop the index i is outside of the array bounds of val. You also need to check the return value of scanf to make sure the input is valid. The last printf statement also needs a trailing newline.
Edit: Note that LEN only handles "real" arrays; arrays passed to functions are received as pointers.
You allocated only int /* ... */ val[99] (only val[0] to val[98] are available) and accessed upto val[100] because the loop condition is i<101.
This will lead to dangerous out-of-range write (undefined behaior).
Allocate enough elements like int /* ... */ val[101] or fix the loop condition not to cause out-of-range access.
Also you didn't set value of i after the for (i=1;i<101;i++) loop, so value of uninitialized element will be used in the do ... while loop. Values of uninitialized elements of non-static local variables are indeterminate and using the value invokes undefned behavior.
Set i to proper value before the loop or change the indice i to proper thing.
Related
I wrote this little code just to start learning some if statements and C coding in general. However, there is an issue. When running it, if the largest element is the last one, the code won't recognize it. Why is that?
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main(){
int num[100];
int max;
int i;
printf("Enter 10 numbers: \n");
for(i = 1; i < 10; i++){
scanf("%d\n", &num[i]);
}
max = num[0];
for(i = 0; i < 10; i++){
if(max < num[i]){
max = num[i];
}
}
printf("The biggest nr is: %d", max);
return 0;
}
Your first loop should start from 0, not 1.
for(i = 0; i < 10; i++){
scanf("%d\n", &num[i]);
}
max already starts with an uninitialized value, here be dragons.
Inside of:
for (i = 1; i < 10; i++) {
scanf("%d\n", &num[i]);
}
max = num[0];
max has an indeterminate value because the loop's counter variable i starts at 0, not 1 which gives the result that the first element of the array wasn't assigned inside of the loop. So you end up assigning this indeterminate value to max.
To use an indeterminate value in the following code:
if (max < num[i]) {
max = num[i];
}
invokes undefined behavior.
"However, if I change i=0, the program asks me for 11 inputs before moving on. And among those 11 inputs, the program still won't count the last one, if it is the largest."
"When running it, if the largest element is the last one, the code won't recognize it. Why is that?"
It doesn't actually ask you for an 11th input for any presumed 11th array element as you think it does and the last in the loops1 treated element of the array is not the one you think it is. That is just an impression to you.
This behavior is caused by the newline character in the format string of the scanf() call:
scanf("%d\n", &num[i]);
The newline character (\n ) is equal to any white space and with this directive, scanf() reads an unlimited amount of white space characters until it finds any-non white space character in the input to stop consuming and the control flow continues to the next statement.
Why does scanf ask twice for input when there's a newline at the end of the format string?
It doesn't ask for the input of the 11th element of the array (as you think it does). It simply needs any non-white space character that the directive fails.
The last element of the array (which is treated inside of the loops1) is still the 10th (num[9]), not the 11th (num[10]) and so is the output correct when you initialize the counter to 0 and it prints:
The biggest nr is: 10
because 10 is the value of the last treated element num[9].
1) Note that you made a typo at the declaration of num -> int num[100];. With this you define an array of one hundred elements, but you actually only need one of 10 elements -> int num[10];.
Side Note:
Also always check the return value of scanf()!
if (scanf("%d\n", &num[i]) != 1)
{
// Error routine.
}
There are two problems in the code one after another:
The loop should begin from 0 instead of 1:
for (int i = 0; i < 10; i++)
The main problem is here:
scanf("%d\n", &num[i]);
_________^^____________
Remove the \n and your problem will be fixed.
I am a beginner to C language and also computer programming. I have been trying to solve small problems to build up my skills. Recently, I am trying to solve a problem that says to take input that will decide the number of series it will have, and add the first and last number of a series. My code is not working and I have tried for hours. Can anyone help me solve it?
Here is what I have tried so far.
#include<stdio.h>
int main()
{
int a[4];
int x, y, z, num;
scanf("%d", &num);
for (x = 1; x <= num; x++) {
scanf("%d", &a[x]);
int add = a[0] + a[4];
printf("%d\n", a[x]);
}
return 0;
}
From from your description it seems clear that you should not care for the numbers in between the first and the last.
Since you want to only add the first and the last you should start by saving the first once you get it from input and then wait for the last number. This means that you don't need an array to save the rest of the numbers since you are not going to use them anyway.
We can make this work even without knowing the length of the series but since it is provided we are going to use it.
#include<stdio.h>
int main()
{
int first, last, num, x = 0;
scanf("%d", &num);
scanf("%d", &first);
last = first; //for the case of num=1
for (x = 1; x < num; x++) {
scanf("%d", &last);
}
int add = first + last;
printf("%d\n", add);
return 0;
}
What happens here is that after we read the value from num we immediately scan for the first number. Afterwards, we scan from the remaining num-1 numbers (notice how the for loop runs from 1 to num-1).
In each iteration we overwrite the "last" number we read and when the for loop finishes that last one in the series will actually be the last we read.
So with this input:
4 1 5 5 1
we get output:
2
Some notes: Notice how I have added a last = first after reading the first number. This is because in the case that num is 1 the for loop will never iterate (and even if it did there wouldn't be anything to read). For this reason, in the case that num is 1 it is reasonably assumed that the first number is also the last.
Also, I noticed some misconceptions on your code:
Remember that arrays in C start at 0 and not 1. So an array declared a[4] has positions a[0], a[1], a[2] and a[3]. Accessing a[4], if it works, will result in undefined behavior (eg. adding a number not in the input).
Worth noting (as pointed in a comment), is the fact that you declare your array for size 4 from the start, so you'll end up pretending the input is 4 numbers regardless of what it actually is. This would make sense only if you already knew the input size would be 4. Since you don't, you should declare it after you read the size.
Moreover, some you tried to add the result inside the for loop. That means you tried to add a[0]+a[3] to your result 4 times, 3 before you read a[3] and one after you read it. The correct way here is of course to try the addition after completing the input for loop (as has been pointed out in the comments).
I kinda get what you mean, and here is my atttempt at doing the task, according to the requirement. Hope this helps:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int first, last, num, x=0;
int add=0;
printf("What is the num value?\n");//num value asked (basically the
index value)
scanf("%d", &num);//value for num is stored
printf("What is the first number?\n");
scanf("%d", &first);
if (num==1)
{
last=first;
}
else
{
for (x=1;x<num;x++)
{
printf("Enter number %d in the sequence:\n", x);
scanf("%d", &last);
}
add=(first+last);
printf("Sum of numbers equals:%d\n", add);
}
return 0;
}
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I am trying to sum an array of numbers. The array has a length determined by an input and then the user gives the array. There were no compilation errors and I am able to run other programs. On the immediate start of running the program I am given a message that program has stopped working and that windows is searching for solution.
#include <stdio.h>
int main()
{
int sum, length, count;
int array[length];
sum=0;
scanf("%d",&length);
scanf("%d",&sum);
for(count=0; count<length-1; count++)
{
sum = sum + array[count];
}
printf("%d", sum);
return 0;
}
When you declare your array it depends on length but you ask the user for length after.
A solution could be to ask the user for length (scanf("%d",&length);) before declaring your actual array (int array[length];).
you should move int array[length] to after scanf("%d", &length). But it is not allowed in C to declare variables after the first non-declaration (it is however possible if you compile this program as C++).
In fact, in standard C you can't have a non-const length definition for an array variable. gcc on the other hand for example allows this nevertheless.
In your case, the problem is that length has an undefined value at the declaration of int array[length];. If you are lucky, your data segment has been initialized to zero (there is no guarantee for that) but otherwise, it may be any value, including a value which leads the program to exceed your physical memory.
A more standard way of doing this is:
int *array = NULL;
scanf("%d",&length);
...
array = (int*) malloc(sizeof(int) * length);
...
free(array);
By the way, even after fixing that, you will most likely get random numbers because you never actually assign the contents of the elements of array.
Local variable are initialized to 0. Hence value of length is 0. So you array is of length. You are then reading length, say 10, from stdin and expect the array to be of length 10. This can't be. Since this is a stack variable, the size is determined in time of pre-processing and not in run time. If you want to define the array length in run time then use malloc.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
int sum, length, count;
int *array;
sum=0;
scanf("%d", &length);
scanf("%d",&sum);
array = (int *)malloc(sizeof(int) * length);
if (array == NULL) return 0;
memset(array, length, 0);
for(count=0; count<length-1; count++)
{
sum = sum + array[count];
}
printf("%d", sum);
return 0;
}
Thanks.
first problem:
the length variable is being used to set the number of entries in the array[], before the variable length is set. Therefore, length will contain what ever trash happens to be on the stack when the program starts so the number of entries defined in array[] is an unknown.
This results in undefined behaviour and could lead to a seg fault event, depending on what was on the stack and what the user entered for length.
second problem:
The array array[] is never initialized so will contain what ever trash is on the stack at program startup. This means the value being printed could be anything. And the 'sum' could overflow, depending on the trash values in array[]
OP program lacks the part of data input, it's asking for sum instead of the values to sum, which is weird. The only inputs requested are also never checked (the return value of scanf must always be checked).
In C (at least C99 and optionally C11) Variable Length Arrays, like the one defined by int array[length], can be used, but the variable length here is used uninitialized and before it is even asked to the user.
Moreover, the loop where the sum is calculated stops before the last element of the array (not really a big deal in this case, considering that all those variables are uninitialized...).
A better way to perform this task could be this:
#include <stdio.h>
// helper function to read an integer from stdin
int read_int( int *value ) {
int ret = 0;
while ( (ret = scanf("%d", value)) != 1 ) {
if ( ret == EOF ) {
printf("Error: Unexpected end of input.\n");
break;
}
scanf("%*[^\n]"); // ignore the rest of the line
printf("Please, enter a number!\n");
}
return ret;
}
int main(void) {
int sum = 0,
length = 0,
count,
i;
printf("Please, enter the number of values you want to add: ");
if ( read_int(&length) == EOF )
return -1;
// Use a VLA to store the numbers
int array[length];
// input the values
for ( count = 0; count < length; ++count ) {
// please, note ^^^^^^^^ the range check
printf("Value n° %2d: ", count + 1);
if ( read_int(&array[count]) == EOF ) {
printf("Warning: You entered only %d values out of %d.\n",
count, length);
break;
}
// you can sum the values right here, without using an array...
}
// sum the values in the array
for ( i = 0; i < count; ++i ) {
// ^^^^^^^^^ sum only the inputted values
sum += array[i];
}
printf("The sum of the values is:\n%d\n", sum);
return 0;
}
I want to add numbers to an array using scanf
What did i do wrong? it says expected an expression on the first bracket { in front of i inside the scanf...
void addScores(int a[],int *counter){
int i=0;
printf("please enter your score..");
scanf_s("%i", a[*c] = {i});
}//end add scores
I suggest:
void addScores(int *a, int count){
int i;
for(i = 0; i < count; i++) {
printf("please enter your score..");
scanf("%d", a+i);
}
}
Usage:
int main() {
int scores[6];
addScores(scores, 6);
}
a+i is not friendly to newcomer.
I suggest
scanf("%d", &a[i]);
Your code suggests that you expect that your array will be dynamically resized; but that's not what happens in C. You have to create an array of the right size upfront. Assuming that you allocated enough memory in your array for all the scores you might want to collect, the following would work:
#include <stdio.h>
int addScores(int *a, int *count) {
return scanf("%d", &a[(*count)++]);
}
int main(void) {
int scores[100];
int sCount = 0;
int sumScore = 0;
printf("enter scores followed by <return>. To finish, type Q\n");
while(addScores(scores, &sCount)>0 && sCount < 100);
printf("total number of scores entered: %d\n", --sCount);
while(sCount >= 0) sumScore += scores[sCount--];
printf("The total score is %d\n", sumScore);
}
A few things to note:
The function addScores doesn't keep track of the total count: that variable is kept in the main program
A simple mechanism for end-of-input: if a letter is entered, scanf will not find a number and return a value of 0
Simple prompts to tell the user what to do are always an essential part of any program - even a simple five-liner.
There are more compact ways of writing certain expressions in the above - but in my experience, clarity ALWAYS trumps cleverness - and the compiler will typically optimize out any apparent redundancy. Thus - don't be afraid of extra parentheses to make sure you will get what you intended.
If you do need to dynamically increase the size of your array, look at realloc. It can be used in conjunction with malloc to create arrays of variable size. But it won't work if your initial array is declared as in the above code snippet.
Testing for a return value (of addScores, and thus effectively of scanf) >0 rather than !=0 catches the case where someone types ctrl-D ("EOF") to terminate input. Thanks #chux for the suggestion!
I am learning array and just wrote this small program to see how it works. but its crashing with segmentation faul which i understand mean i am writing my variable / function to an memory place not allotted to it. But I cant figure how. Can anyone let me know please?
i am calling introArray from my main().
int introArray (void)
{
int total, ctr;
printf("enter how many students \n");
scanf("%d", &total);
int students[total];
ctr = 0;
while ( students[ctr] <= total)
{
printf("enter student %d DOB in mmddyy \n", ctr );
scanf("%d", students[ctr]);
ctr++;
}
return 0;
}
In your code, there is one implementation logic issue. The total number of students is total and hence, your while loop should be
while(ctr < total)
The data to be read also should scanf("%d", &students[ctr]); There is an ampersand missing
ctr goes beyond total. This way you are going out of bound
Change the loop to
while (ctr < total)
{
printf("enter student %d DOB in mmddyy \n", ctr );
scanf("%d", &(students[ctr]));
ctr++;
}
I think
scanf("%d", students[ctr]);
should be
scanf("%d", &students[ctr]);
This line
while ( students[ctr] <= total)
is not protection against reading past your array bounds inside the loop. This will stop you reading past the end of your array provided you use ctr as your index
while ( ctr < total)
you need the strict inequality as array indices are zero based.
In addition, your scanf call inside your while loop is wrong - the second argument should be a pointer and currently you pass an integer. It should be
scanf("%d", &students[ctr]);