I am breaking my head on this but cannot proceed so please help.
Working on a programming assignment:
INPUT:
First line contains a value N representing the dimension of the input matrix, followed by N lines, each line representing a row of the matrix. Within each row, N values are given and are separated by whitespace.
So instead of using fgets() and then extracting the digits using strtol() I am forced to use scanf() because the assignment platform does not work properly with strtol().
This is my program:
int main()
{
int N;
int arr[N][N];
int idx1, idx2=0;
scanf("%d ",&N);
for (idx1=0; idx1<N; idx1++) {
for(idx2=0; idx2<N; idx2++) {
scanf("%d", &arr[idx1][idx2]);
printf("arr[%d][%d]=%d ", idx1, idx2, arr[idx1][idx2]);
}
}
printf("\n");
for (idx1=0; idx1<N; idx1++) {
for(idx2=0; idx2<N; idx2++) {
printf("arr[%d][%d]=%d ", idx1, idx2, arr[idx1][idx2]);
}
}
return 0;
}
Output:
3
1 2 3
arr[0][0]=1 arr[0][1]=2 arr[0][2]=3
4 5 6
arr[1][0]=4 arr[1][1]=5 arr[1][2]=6
7 8 9
arr[2][0]=7 arr[2][1]=8 arr[2][2]=9
arr[0][0]=1 arr[0][1]=4 arr[0][2]=7 arr[1][0]=4 arr[1][1]=7 arr[1][2]=8 arr[2][0]=7 arr[2][1]=8 arr[2][2]=9
FWIW gcc --version
gcc (Ubuntu/Linaro 4.6.3-1ubuntu5) 4.6.3
Why is the second output not matching the first? What am i missing here?
int N;
int arr[N][N];
int idx1, idx2=0;
scanf("%d ",&N);
You have to declare arr after you write N object. Here when arr is declared, N value is indeterminate.
If you want to create a variable length array like that, you need to dynamically allocate it instead of creating an array of size N (N has some garbage value at the time) and then changing the size of it based on user input. Use malloc or calloc.
Related
I'm trying to create a complete C program to read ten alphabets and display them on the screen. I shall also have to find the number of a certain element and print it on the screen.
#include <stdio.h>
#include <conio.h>
void listAlpha( char ch)
{
printf(" %c", ch);
}
int readAlpha(){
char arr[10];
int count = 1, iterator = 0;
for(int iterator=0; iterator<10; iterator++){
printf("\nAlphabet %d:", count);
scanf(" %c", &arr[iterator]);
count++;
}
printf("-----------------------------------------");
printf("List of alphabets: ");
for (int x=0; x<10; x++)
{
/* I’m passing each element one by one using subscript*/
listAlpha(arr[x]);
}
printf("%c",arr);
return 0;
}
int findTotal(){
}
int main(){
readAlpha();
}
The code should be added in the findTotal() element. The output is expected as below.
Output:
List of alphabets : C C C A B C B A C C //I've worked out this part.
Total alphabet A: 2
Total alphabet B: 2
Total alphabet C: 6
Alphabet with highest hit is C
I use an array to count the number of the existence of each character,
I did this code but the display of number of each character is repeated in the loop
int main()
{
char arr[100];
printf("Give a text :");
gets(arr);
int k=strlen(arr);
for(int iterator=0; iterator<k; iterator++)
{
printf("[%c]",arr[iterator]);
}
int T[k];
for(int i=0;i<k;i++)
{
T[i]=arr[i];
}
int cpt1=0;
char d;
for(int i=0;i<k;i++)
{int cpt=0;
for(int j=0;j<k;j++)
{
if(T[i]==T[j])
{
cpt++;
}
}
if(cpt>cpt1)
{
cpt1=cpt;
d=T[i];
}
printf("\nTotal alphabet %c : %d \n",T[i],cpt);
}
printf("\nAlphabet with highest hit is : %c\n",d,cpt1);
}
There is no way to get the number of elements You write in an array.
Array in C is just a space in the memory.
C does not know what elements are actual data.
But there are common ways to solve this problem in C:
as mentioned above, create an array with one extra element and, fill the element after the last actual element with zero ('\0'). Zero means the end of the actual data. It is right if you do not wish to use '\0' among characters to be processed. It is similar to null-terminated strings in C.
add the variable to store the number of elements in an array. It is similar to Pascal-strings.
#include <stdio.h>
#include <string.h>
#define ARRAY_SIZE 10
char array[ARRAY_SIZE + 1];
int array_len(char * inp_arr) {
int ret_val = 0;
while (inp_arr[ret_val] != '\0')
++ret_val;
return ret_val;
}
float array_with_level[ARRAY_SIZE];
int array_with_level_level;
int main() {
array[0] = '\0';
memcpy(array, "hello!\0", 7); // 7'th element is 0
printf("array with 0 at the end\n");
printf("%s, length is %d\n", array, array_len(array));
array_with_level_level = 0;
const int fill_level = 5;
int iter;
for (iter = 0; iter < fill_level; ++iter) {
array_with_level[iter] = iter*iter/2.0;
}
array_with_level_level = iter;
printf("array with length in the dedicated variable\n");
for (int i1 = 0; i1 < array_with_level_level; ++i1)
printf("%02d:%02.2f ", i1, array_with_level[i1]);
printf(", length is %d", array_with_level_level);
return 0;
}
<conio.h> is a non-standard header. I assume you're using Turbo C/C++ because it's part of your course. Turbo C/C++ is a terrible implementation (in 2020) and the only known reason to use it is because your lecturer made you!
However everything you actually use here is standard. I believe you can remove it.
printf("%c",arr); doesn't make sense. arr will be passed as a pointer (to the first character in the array) but %c expects a character value. I'm not sure what you want that line to do but it doesn't look useful - you've listed the array in the for-loop.
I suggest you remove it. If you do don't worry about a \0. You only need that if you want to treat arr as a string but in the code you're handling it quite validly as an array of 10 characters without calling any functions that expect a string. That's when it needs to contain a 0 terminator.
Also add return 0; to the end of main(). It means 'execution successful' and is required to be conformant.
With those 3 changes an input of ABCDEFGHIJ produces:
Alphabet 1:
Alphabet 2:
Alphabet 3:
Alphabet 4:
Alphabet 5:
Alphabet 6:
Alphabet 7:
Alphabet 8:
Alphabet 9:
Alphabet 10:-----------------------------------------List of alphabets: A B C D E F G H I J
It's not pretty but that's what you asked for and it at least shows you've successfully read in the letters. You may want to tidy it up...
Remove printf("\nAlphabet %d:", count); and insert printf("\nAlphabet %d: %c", count,arr[iterator]); after scanf(" %c", &arr[iterator]);.
Put a newline before and after the line of minus signs (printf("\n-----------------------------------------\n"); and it looks better to me.
But that's just cosmetics. It's up to you.
There's a number of ways to find the most frequent character. But at this level I recommend a simple nested loop.
Here's a function that finds the most common character (rather than the count of the most common character) and if there's a tie (two characters with the same count) it returns the one that appears first.
char findCommonest(const char* arr){
char commonest='#'; //Arbitrary Bad value!
int high_count=0;
for(int ch=0;ch<10;++ch){
const char counting=arr[ch];
int count=0;
for(int c=0;c<10;++c){
if(arr[c]==counting){
++count;
}
}
if(count>high_count){
high_count=count;
commonest=counting;
}
}
return commonest;
}
It's not very efficient and you might like to put some printfs in to see why!
But I think it's at your level of expertise to understand. Eventually.
Here's a version that unit-tests that function. Never write code without a unit test battery of some kind. It might look like chore but it'll help debug your code.
https://ideone.com/DVy7Cn
Footnote: I've made minimal changes to your code. There's comments with some good advice that you shouldn't hardcode the array size as 10 and certainly not litter the code with that value (e.g. #define ALPHABET_LIST_SIZE (10) at the top).
I have used const but that may be something you haven't yet met. If you don't understand it and don't want to learn it, remove it.
The terms of your course will forbid plagiarism. You may not cut and paste my code into yours. You are obliged to understand the ideas and implement it yourself. My code is very inefficient. You might want to do something about that!
The only run-time problem I see in your code is this statement:
printf("%c",arr);
Is wrong. At this point in your program, arr is an array of char, not a single char as expected by the format specifier %c. For this to work, the printf() needs to be expanded to:
printf("%c%c%c%c%c%c%c%c%c%c\n",
arr[0],arr[1],arr[2],arr[3],arr[4],
arr[5],arr[6],arr[7],arr[8],arr[9]);
Or: treat arr as a string rather than just a char array. Declare arr as `char arr[11] = {0};//extra space for null termination
printf("%s\n", arr);//to print the string
Regarding this part of your stated objective:
"I shall also have to find the number of a certain element and print it on the screen. I'm new to this. Please help me out."
The steps below are offered to modify the following work
int findTotal(){
}
Change prototype to:
int FindTotal(char *arr);
count each occurrence of unique element in array (How to reference)
Adapt above reference to use printf and formatting to match your stated output. (How to reference)
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;
}
Given an array of integers print each element in reverse order as a single line of space-separated integers.
Input Format
Output Format
Print all integers in in reverse order as a single line of space-separated integers.
Sample Input
4
1 4 3 2
Sample Output
2 3 4 1
Here's my Code
int main()
{
int n;
int a[n];
cin >>n;
for (int i=0;i<n;i++)
{
cin>>a[i];
}
for (int i=n-1;i>=0;i--)
cout<<a[i]<<" ";
return 0;
}
You can not allocate an static array with a variable size meaning the expression a[n] is not acceptable since in the compile time the variable "n" is not determined. If this question have a limit for "n" - for example n < 1000 - then replace the expression with int a[1000] ;. If it doesn't , replace it with int* a = new int[n] ;. It is called dynamic allocation , you can search it to find out how amazing it is ;)
I am new in c programming and I am studying arrays right now. I saw that unlike java you can't print them automatically by using Arrays.toString() method. I want to write a simple function that prints the array.
In the program it asks us the size of the array and when we write it, it asks what is the value for each element and then the program calls the displayArray() function to print the array on a single line.
For example :
Hello. What will be the size of the new array?
3
Enter the 1. element
7
Enter the 2. element
5
Enter the 3. element
1
The result should be "Your array is: 7 5 1" but instead of that I get "Your array is: 3 6356728 2" as a result. Can you help?
#include <stdio.h>
void displayArray();
int main()
{
int size;
printf("Hello. What will be the size of the new array?\n");
scanf("%d", &size);
int myarray[size];
for (int i = 0; i < size; i++)
{
printf("Enter the %d. element\n" , (i + 1));
scanf("%d", &myarray[i]);
}
displayArray(myarray[size], size);
return 0;
}
void displayArray(int myarray[], int size)
{
printf("Your array is: ");
for (int i = 0; i < size; i++)
{
printf("%d ", myarray[i]);
}
return;
}
You have a problem in the function call (which your compiler should have warned you about, read the footnote later)
displayArray(myarray[size], size);
should be
displayArray(myarray, size);
because,
Type mismatch between formal parameter and actual argument: you function expects an array (a pointer to the first element of the array, to be precise), not an element of the array.
undefined behavior. For an array defined as int myarray[size], accessing element like myarray[size] is off-by-one, as C arrays are 0-based indexing.
Footnote:
If you try to compile your code, compiler should complain about the mismatch. It can happen either
You did not turn up the compiler warning level (which is a mistake from your side)
or, you chose to ignore the warnings (which is an "offense" from your side)
It's a common mistake which beginners commit.
While calling a function in which you pass an array, you only need to specify the name of the array you want to pass.
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!