I have as input integer,cahr,symbol random array. I want to extract only integers from that array.
For example: input is [#56gY68#$&*+7j^78gu5];
output is 56 68 7 78 5
Here is my code:
#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
#include<ctype.h>
void main()
{
char arr[10];
int arr1[10],i;
clrscr();
for(i=0;i<10;i++)
{
scanf("%s",arr[i]);
}
for(i=0;i<10;i++){
if(isdigit(arr[i])){
arr1[i]=arr[i];
}
}
for(i=0;i<10;i++){
printf("%d",arr1[i]);
}
getch();
}
The code fails to extract an integer. What do I need to change?
The code shown here invokes undefined behaviour. The problems in your code:
scanf() requires a pointer type argument to store the scanned value.
%s is for strings, %c is for char.
for %c, the \n is not ignored. So, you've to use it like " %c" to avoid the trailing newline.
However, I don't see any logic to "extract only one integer from a random array". You may need to check your logic once again.
Notes:
The recommended signature of main() is int main(void).
always initalize all your local variables.
Related
I'm running a while loop so the user can constantly enter expressions, until they indicate they want to quit the program. I'm using strcmp() to compare two strings so as soon as they enter quit the program will stop. But the program keeps going, any Ideas?
#include <stdio.h>
#include <string.h>
int main()
{
int min12=0;
char opper;
int x=0;
int min13;
char *Repeatprog="cont";
char *Repeatprog1="quit";
while (strcmp(Repeatprog,Repeatprog1))
{
printf("enter the integer number \n");
scanf( "%d %c %d", &min12, &opper, &min13);
printf("%d %c %d\n", min12, opper, min13);
printf("Type the word quit to end program\n");
scanf("%s", Repeatprog);
}
printf("Good Bye");
return 0;
}
Remember always that an Array is a Pointer to the first object of the array.
And secondly, in your call to scanf() you only read a character. Not a whole string (represented by %s in C)
So in conclusion, your call to scanf() shouldn't have a pointer and should have a string instead of a character.
scanf("%s", Repeatprog);
or simply
gets (Repeatprog);
EDIT :
As the commenter #EOF said, gets() is not a good idea since it can lead to Undefined Behaviour. That's because the program can read more characters than it should have and lead to overflow, thus it isn't secure.
So I recommend using char *fgets(char *str, int n, FILE *stream)
Note:
Also, your code is using string literals. So if you make any attempt to change the content of the char pointer then it will lead to Undefined Behaviour.
For this note, please thank the guys below me [comments]. I made a huge mistake and I'm sorry.
i am starting to learn C and I have ran into the problem of adding a string input to a 2D array, i am able to get the string input correctly but when i try and add the string element to the array it is not working as expected.When printing the array(which is how i test the program) it will assign each index in the array a single character instead of the entire string.
And here is my code for viewing, thank you very much in advance i appreciate any help that is posted.
#include <stdio.h>
main()
{
char c_array[3][3];
int x;
int y;
int z=10;
int i_user_input;
char c_name[256];
for(x=0;x<=+2;x++)
{
for(y=0;y<=2;y++)
{
printf("\nPlease enter a name to the system:");
scanf(" %s",&c_array[x][y]);
printf("The string is %s\n",c_name);
printf("Please press '1' if you would like to keep entering names\n");
printf("Please press '2' if you would like to print the list of names:");
scanf("%d",&i_user_input);
if(i_user_input==1)
continue;
if(i_user_input==2)
for(x=0;x<=2;x++)
{
for(y=0;y<=2;y++)
{
printf("c_array[%d][%d]=%c\n",x,y,c_array[x][y]);
}
}
}
}
}
The output looks as follows with the sample input 'Penelope!'
c_array[0][0]=P
c_array[0][1]=e
c_array[0][2]=n
c_array[1][0]=e
c_array[1][1]=l
c_array[1][2]=o
c_array[2][0]=p
c_array[2][1]=e
c_array[2][2]=!
When you declare:
char c_array[3][3];
This means that each element of your 2D array is a "char" (a single character); not a "string of characters".
To have each element be a string of characters, you need to declare your array in the following way:
char string_array[3][3][256];
I am not sure this is what you want to do though. Mw feeling is that you want an array of 3 elements where each element is a string. In that case, [3] is hardly enough, your strings will have to be less than two characters (the third being reserved for the terminating zero).
Strings aren't a type. They're a value pattern, like if you say an int stores a multiple of ten, then it ends in 0... If you say an array of char stores a string, then it ends at the first '\0', see? We can store multiples of ten in different kinds of integer variables, and likewise for strings.
Strings are patterns of values, not types. When choosing which integer type we want to use, we consider the range for integers. We choose int for small integer values (less than 32768), long for larger values (less than 2147483648) or long long for values larger than that. As a result, we choose the correct type of integer depending on the multiple of ten we expect. Likewise for strings, we need to make sure we have enough memory for the characters in the string followed by the '\0' at the end.
The character &c_array[x][y] only has enough memory for 0 characters followed by a '\0' at the end; it's only useful for storing an empty string. Perhaps you meant to declare c_array like this:
char c_array[3][3][256];
In this case, scanf expects %s to correspond to an argument of the type char *... but as your compiler will probably warn you, &c_array[x][y] will have the type char (*)[256]. If you want to fix that warning, you'll need to remove the & from your scanf code:
if (scanf("%s", c_array[x][y]) != 1) {
/* XXX: Handle an error here, *
* e.g. by exiting or returning or something */
}
While we're on that topic, you'll notice that I removed the redundant space; %s already performs the functionality that the format directive would perform. I also wrote code to check the return value of scanf, as you should...
Remember how we spoke about choosing the types of integers that we use to store data? One other consideration is whether or not the data we intend to store can be negative. Consider this: In your code, should x and y be able to store negative values? What sense does a negative array index make? It is advisable that all array indexes be declared as size_t. When you use printf to print a size_t value, you'll want to use the %zu format specifier rather than %d.
char c_name[256];
/* ... */
printf("The string is %s\n",c_name);
Now, if the string is stored into c_array[x][y], then what is the point of c_name? That's an unnecessary variable. Use c_array[x][y], instead.
printf("\nPlease enter a name to the system:");
Common implementations will often refrain from printing characters until a '\n' is written; the line is printed all at once rather than character by character. As a result, you might see some strange behaviour such as this line not appearing at the right time. Fix this problem by putting the '\n' on the end of the line, rather than the beginning:
printf("Please enter a name to the system:\n");
... or use puts, which adds a '\n' for you automatically:
puts("Please enter a name to the system:");
... or use fflush, which will write all pending unwritten data even if there is no '\n':
fflush(stdout);
In your code
scanf(" %s",&c_array[x][y]);
printf("The string is %s\n",c_name);
both the statements are wrong.
%s expects a pointer to an array, not a single char. To scan a single char, you need %c format specifier.
c_name, as used in the printf() is uninitialised. Using uninitialised value invokes undefined behaviour.
Solution: To take input element by element, you can do do something like
for(x=0; x<=2 ;x++)
{
printf("Start entering the name, one character at a time\n");
for(y=0; y< 2; y++) //note only '<' here
{
scanf(" %c", &c_array[x][y]); // note the leading space
}
c_array[x][y] = 0; //null termination of array to be used as string
}
If you want to assign a string to each index in the array, then create the array as follows :
#include<stdio.h>
int main(void)
{
char a[2][10];
int i=0;
char temp[20];
for(i=0;i<2;i++)
{
scanf("%s",temp);
strcpy(a[i],temp);
}
printf("%s",a[0]);
return 0;
}
Each string that you enter will be stored in each index of the array.
#include <stdio.h>
#include <string.h>
int main(void){
char c_array[3][3];
int x;
int y;
//int z=10;//unused
int i_user_input;
char c_name[256];
printf("\nPlease enter a name to the system:");
scanf("%255s",c_name);
printf("The string is %s\n",c_name);
printf("Please press '1' if you would like to keep entering names\n");
printf("Please press '2' if you would like to print the list of names:");
scanf("%d", &i_user_input);
if(i_user_input==1)
;
else if(i_user_input==2){
memcpy(c_array, c_name, sizeof(c_array));
for(x=0;x<3;x++){
for(y=0;y<3;y++){
printf("c_array[%d][%d]=%c\n", x, y, c_array[x][y]);
}
}
}
}
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define SIZE 3
int main(void){
char *c_array[SIZE][SIZE] = { NULL };
int x;
int y;
//int z=10;//unused
int i_user_input;
char c_name[256];
for(x=0;x<SIZE;x++){
for(y=0;y<SIZE;y++){
printf("\nPlease enter a name to the system:");
scanf("%255s", c_name);
printf("The string is %s\n", c_name);
printf("Please press '1' if you would like to keep entering names\n");
printf("Please press '2' if you would like to print the list of names:");
scanf("%d", &i_user_input);
c_array[x][y] = strdup(c_name);
if(i_user_input==1){
continue;
}
if(i_user_input==2){
int x, y;//local variable
for(x=0;x<SIZE;x++){
for(y=0;y<SIZE;y++){
if(c_array[x][y] != NULL)
printf("c_array[%d][%d]=%s\n", x, y, c_array[x][y]);
}
}
}
}
}
for(x=0;x<SIZE;x++)
for(y=0;y<SIZE;y++)
free(c_array[x][y]);
}
Why not use pointers in you array/matrix?
#include <stdio.h>
#include <string.h>
main()
{
char *c_array[3][3];
int x;
int y;
int z=10;
int i_user_input;
char c_name[256];
for(x=0;x<=+2;x++)
{
for(y=0;y<=2;y++)
{
printf("\nPlease enter a name to the system:");
scanf(" %s",c_name);
new_string = malloc(strlen(c_name)+1)
if (new_string == NULL) {
print("error allocating string\n")
exit(-1);
}
strcpy(new_string, c_name);
c_array[x][y] = new_string
printf("The string is %s\n",c_name);
printf("Please press '1' if you would like to keep entering names\n");
printf("Please press '2' if you would like to print the list of names:");
scanf("%d",&i_user_input);
if(i_user_input==1)
continue;
if(i_user_input==2)
for(x=0;x<=2;x++)
{
for(y=0;y<=2;y++)
{
printf("c_array[%d][%d]=%c\n",x,y,c_array[x][y]);
}
}
}
}
for(x=0;x<=2;x++) {
for(y=0;y<=2;y++) {
free(c_array[x][y])
}
}
}
The code above stores the input string in each cell of the array, if that is your goal.
Note: I have no tried it, so there may be subtle errors. My point is to show you how to use pointers and their equivalence to arrays. Using C without pointers is pointless. :)
I'm trying to write a short program were:
#include <stdio.h>
void main()
{
char=a,b,c;
printf("please place 3 numbers:\n");
scanf("%c%c%c", &a,&b,&c);
}
The exercise I'm trying to solve is how to change the char to int so if I write in a the number 3, I will get the number 3 Printed.
at this point I'm only getting the value.
I would appreciate any help.
The answer depends somewhat on what you can assume about the character set. If it's something like ASCII (or really, any character set that includes the digits in sequential order), you just need to offset the character value by the value of the character 0:
int aValue = a - '0';
I'm sure that C# provides better ways to do what you're trying to do, though. For example, see this question for some examples of converting strings to integer values.
First of all your syntax need some checking
You should know that you declare a variable this way (a char in this example):
char a;
If you want to declare multiple variables of the same type in a row you do :
char a, b, c;
If you want to assign a value to a declared variable :
a = '3';
Now to print a char using printf (man printf is a must read, more infos are in coreutils) :
printf("%c", a);
If you want to get the char from the command line, I recommand you to use getchar() (man getchar) instead of scanf because if suits better what you are trying to achieve and doesn't require you to use a syntax in scanf that I am sure you don't fully understand yet.
Your question is incredibly light on details, so here are several options:
#include <stdio.h>
int main()
{
char a,b,c;
printf("please place 3 numbers:\n");
scanf("%c%c%c", &a,&b,&c);
printf("Printing ints (auto-promotion): %d %d %d\n", a, b, c);
printf("Printing ints (explicit-promotion): %d %d %d\n", (int)a, (int)b, (int)c);
printf("Printing digits: %d %d %d\n", a-0x30, b-0x30, c-0x30);
return 0;
}
If the input is 123,
I expect the output to be:
Printing ints (auto-promotion): 49 50 51
Printing ints (explicit-promotion): 49 50 51
Printing digits: 1 2 3
Some things I fixed along the way.
main should return an int, not be void.
char=a,b,c; is a syntax error. You meant char a,b,c;
added a return 0; at the end of main.
You question is not quite understandable. Still I'll try to help. I think that what you want is to store an integer value in the char variable. You can do so by using the following code:
#include<stdio.h>
void main()
{
char a,b,c;
printf("Enter three numbers:\n");
scanf(" %c %c %c",&a,&b,&c); //notice the spaces between %c
}
Or if you want to enter a character and print its ASCII value, you can use the following code:
#include<stdio.h>
#include<conio.h>
void main()
{
char a,b,c;
printf("Enter three characters:\n");
scanf(" %c %c %c",&a,&b,&c);
printf("Entered values: %d %d %d",a,b,c);
getch();
}
My code only takes 5 values as input.? What am i doing wrong?
#include<stdio.h>
#include<stdlib.h>
int main()
{
char arr[3][3];
int i,j,n;
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
{
scanf("%c",&arr[i][j]);
}
}
return 0;
}
How should i correct it?
Change
scanf("%c",&arr[i][j]);
to
scanf(" %c",&arr[i][j]);.
Notice the space given before specifier to consume \n left in stdin buffer when you pressed enter.
Each \n is working as input taking your space from input space.
It should work, but note that %c will read only a single character. Whitespace is not suppressed as it is for other (more "high-level") format specifiers, so if you're separating your characters by blanks of any kind those will be read instead of the actual characters.
Also note that you should check the return value of scanf(), it can fail if no suitable input is present.
The first input is stored in arr[0][0] then when you press enter(return key) it is stored in arr[0][1] and while you think you are inputting the second character actually you are giving the third input. Try the code below to see if you didnt get it.
#include<stdio.h>
#include<stdlib.h>
int main()
{
char arr[3][3];
int i,j,n;
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
{
printf("Input array [%d][%d]",i,j);
scanf("%c",&arr[i][j]);
}
}
}
And as for the correction you need a scanf(" %c",&arr[i][j]); a space infront of %c to consume the \n
Hope it answers your question
Instead you can use gets(arr[i][j])
instead of scanf("%c",&arr[i][j]);
This will work perfectly. It is good idea to use gets() and puts() function instead of printf() and scanf() function for string and character in C
you can use fflush(stdin)
char arr[3][3];
int i,j,n;
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
{
scanf("%c",&arr[i][j]);
fflush(stdin);
}
}
I am currently trying to solve a problem from CodeChef but I am having troubles with using fgets() inside a loop.
The first input (T) is going to be a positive integer containing the number of user inputs.
Then delimited by newline characters, the user is going to input a string below the length of 10 under any circumstances.
So, I've tried this:
#include <stdio.h>
#include <stdlib.h>
#define SIZE 10
int main()
{
int T;
int diffX, diffY;
char s[SIZE];
scanf("%d", &T);
while (T--){
fgets(s, SIZE, stdin);
printf("%s\n", s);
}
return 0;
}
However, when I attempted to test the code with the following inputs:
3 Hello Hi What
I was only able to input until "Hi" then the program exited successfully (returning 0).
Why is this the case and how can I fix it?
Thank you in advance,
kpark.
fgets() consumes the newline left behind by the first call to scanf(). So, it is consuming 3 lines, but the first line looks like an empty line to the fgets() loop you have.
You can fix this by using fgets() to get the first line too, and parse the string into a number using sscanf().
fgets(s, SIZE, stdin);
sscanf(s, "%d", &T);
/* ... */
It is counting the read of the T as part of the counting. Add a newline in the scanf.
#include <stdio.h>
#include <stdlib.h>
#define SIZE 10
int main()
{
int T;
int diffX, diffY;
char s[SIZE];
scanf("%d\n", &T);
while (T--){
fgets(s, SIZE, stdin);
printf("%s\n", s);
}
return 0;
}
Is your Question is about how to read Multiple Strings in C ?
Then it can be done by 2 ways :-
1.By declaring two dimensional Array of characters.
//Let say we want 6 strings each of them having max 10 characters.
char set[6][10] ;
for(int i=0;i<6;i++)
scanf("%s",set[i])
2.By declaring one dimensional Array of pointers to character (Notice the naming Conventions), in which each of those pointer pointing to a String.
int main(){
int i,numOfStrings;
char temp[30];
printf("Enter Number of strings in set ");
scanf("%d",&numOfStrings);
//Here We have defined array of pointer that will store each string sepratly.
//Arry of pointer to character.
char *setOfStrings[numOfStrings];
for(i=0;i<numOfStrings;i++)
{
printf("Enter string ");
scanf("%s",temp);
setOfStrings[i]= (char*)malloc(sizeof(temp)); //allocted new memory and gave it to array of pointer
strcpy(setOfStrings[i],temp);
}
for(i=0;i<numOfStrings;i++)
{
printf("string = %s \n",setOfStrings[i]);
}
return 0;
}
But that need to understand :
In case of array of pointers we may initialize them with String but Can't take as input from Command line like
char *set[2]={"Dinesh","Kandpal"}; //Its valid but you can't do this from command line
for doing so What we do we will create an space dynamically ,store that address in the one of the element in 1-D array of pointers and then whatever value we have scanned copy that content to the another string to the location that we created using malloc