I am doing homework to school. Task is to make program in C that can do some simple things with matrixes like *, -, +.
My program is reading matrixes and save them to int array. When the matrix input ended, I scanf to char to know what operation iam going to do with the matrixes. Operation are stores in char array. (on input can be up to 100 matrixes so 99 operations in between). If user writes * everything is okay, BUT when user writes '-' or '+' it stores '/n'. I was googling and debuging but still dont know why this happens. Any suggustions are welcome.
User input looks like this: First line is size of matrix.
2 3
76 98 -31
30 30 32
-
2 3
89 25 38
1 -32 -38
int* readMatrix(int width, int height, char arrayOperation[], int numberOfMatrix, int*err)
{
int sizeOfMatrix = ((height*width) + 3), i = 0;
char operation = 'n';
int *arrayMatrix = malloc(sizeOfMatrix* sizeof(int));
arrayMatrix[0] = width;
arrayMatrix[1] = height;
for (i = 2; i <= sizeOfMatrix; i++)
{
if (i == sizeOfMatrix)
{
if (scanf_s("%c", &operation) != EOF)
{
if (operation == '*' || operation == '-' || operation == '+')
{
arrayOperation[numberOfMatrix] = operation;
}
else
{
*err = 100;
}
}
else
{
arrayOperation[numberOfMatrix] = 'n';
break;
}
}
else if (scanf_s("%d", &(arrayMatrix[i])) == 1)
{
*err = 0;
}
}
return arrayMatrix;
}
In case of * the var operation==*,but in -,+ operation==/n. I call this function in loop, output array is stored in array of int**.
scanf("%c" reads the very next character without skipping any whitespace, so it will store whatever is after the last thing read -- usually a newline. If you want to skip whitespace, add a space to the format:
scanf(" %c", &operation)
this will skip any whitespace and store the next non-whitespace character.
You don't need this with %d as all of the conversion patterns except %c and %[ skip whitespace.
I was discusing this particular problem with my lecturer.
I understand this: when scanf reading the number it jumps over whitespace and /n.
So reading "32/n - " results in reading everything. When reading "32/n * " it stops at "*" because its not sign of any number. So its behave little stupid.
This can solve using fgets() function and read it by char.
Related
Hi I want to write a program which takes a string and a number as input and then shifts the elements of string by given number, and also the string is always capital letters and the output should only be capital letters too
example 1 : -1 , "AB CD"
output 1 : "ZA BC"
example 2 : +3 , "ABC"
output 2 : "DEF"
I have written this code but if there is a space in my string , the program doesnt work properly
for example : -1 "AA AA"
the output is : "ZZ" But i was expecting "ZZ ZZ"
int main()
{
char str[100] = {NULL},y;
int n,x,i=0;
scanf_s("%d", &n);
scanf_s("%s", str);
while (str[i] != NULL) {
x = str[i];
if (x + n >= 65 && x + n <= 90) {
y = x + n;
str[i] = y;
}
else if (x + n < 65) {
while (x + n < 65) {
x += 26;
}
y = x + n;
str[i] = y;
}
else if (x + n > 90) {
while (x + n > 90) {
x -= 26;
}
y = x + n;
str[i] = y;
}
i++;
}
printf("%s", str);
return 0;
}
After reading non-whitespace, the scanf specifier %s stops when it encounters whitespace.
Instead of "%s", use " %99[^\n]" to:
read and ignore leading whitespace, then
read up to 99 characters, or until a newline character is encountered.
Reading at most 99 characters is important, as you must leave room in your buffer for the null-terminating byte, and not limiting the amount of data read can easily lead to buffer overflows.
The return value of scanf_s should be checked to ensure the expected number of conversions took place, before any data is used.
if (1 != scanf_s("%d", &n))
/* error! */;
Note that scanf_s requires a secondary argument when using %s or %[, which is the size of the buffer
char buf[10];
if (1 != scanf_s("%9s", buf, (rsize_t) sizeof buf))
/* error! */;
which is supposed to be of type rsize_t. If you are using the MSVC versions of these functions, this type is unsigned.
Due to the pitfalls involved with using scanf properly, and the portability issues of scanf_s, consider instead using fgets to read lines of input (and strtol to parse an integer).
For clarity, consider using character constants like 'A' and 'Z', instead of integer constants like 65 and 90.
As is, your current code does not account for non-alphabetic characters, and will shift those. Consider the use of isalpha or isupper to filter the application of the shift.
Note NULL is usually considered a pointer value. Use '\0', or just 0, to represent the null-terminating byte.
I'm very new to programming and C. I have textfile with some random text and an integer that i want to find and save. The textfile looks something like this (I only want to save 23 in this case, not 56):
# this is a comment
23
this is some random text
56
And this is the code I have:
int *num = malloc(sizeof(int));
*num = fgetc(f);
while(!feof(f)){
*num = fgetc(f);
if(isdigit(*num)){
break;
}
}
if(!isdigit(*num)){
printf("Error: no number found.\n");
free(num);
}
else{
printf("%c\n", *num);
}
I'm kinda stuck right now and my program only prints out the number 2 :/
Very thankful for help.
As #pbn said you're better off using sscanf.
But if you really, really want, you can do it your way, by reading one character at a time, but you'll need to "build" the integer yourself, converting the character to integer, keeping track of what you have, and multiplying by powers of 10 for every digit the number that you already have.
Something like this (not complete code, it's just to get you started):
int c;
int num = 0;
while (c = fgetc(f)) {
if(!isdigit(c)) {
break;
}
num = (num * 10) + (c - '0');
}
The c- '0' part is to convert the text representation of the integer to the integer itself. 0 is character 48, 1 is 49 and so on.
This is assuming that on the line with numbers, you ONLY have numbers, not a mix of numerical and non-numerical characters.
Also, do not use !feof(file).
One option could be using getline and sscanf functions. I assumed that text lines do not contain numbers:
#include <stdio.h>
int main() {
int value, matched = 0;
char *line = NULL;
size_t size;
while(getline(&line, &size, stdin) != -1) {
if ((matched = sscanf(line, "%d", &value)) != 0)
break;
}
if (matched)
printf("value: %d\n", value);
return 0;
}
This part:
while(getline(&line, &size, stdin) != -1) {
will try to read the entire stream line by line.
Next line uses sscanf return value, which is the number of input items successfully matched and assigned, to determine whether the integer value has been found. If so it stops reading the stream.
One simple way in your program is once you find digit don't just stop continue untill you find next " " , "\n" , "\0" . Till then add Number = Number*10 +(*num);, define Number as global or something.
According to the question, The user needs to enter the no of hours the vehicle is parked and the total charge for the hours should get printed beside it.
for example:
I created this simple program
#include<stdio.h>>
#include<math.h>
float calculateCharges(float hurs);
int main()
{
float hours;//total no of hours vehicle is parked
int i;
printf("%s%10s%10s", "Car", "Hours", "Charges");
for (i = 1; i <= 3; i++)
{
printf("\n%d\t", i);
scanf("%f", &hours);
printf("\t%f\n", calculateCharges(hours));
}
getch();
return 0;
}
float calculateCharges(float hurs)
{
float charges;
hurs = ceil(hurs);
if (hurs >= 24) charges = 10;
else
{
if (hurs <= 3) charges = 2;
else
{
hurs = hurs - 3;
charges = 2 + 0.5*hurs;
}
}
return charges;
}
But now every time I enter hours the charges are getting printed below it instead of beside it. As shown in the image:
Is there is a way to consume the newline after scanf? So that charges can be printed beside the scanf?
I have modified my code this way too, but it didn't make any difference.
printf("%s%10s%10s", "Car", "Hours", "Charges");
for (i = 1; i <= 3; i++)
{
printf("\n%d\t", i);
printf("\t%f\n",(scanf("%f", &hours),calculateCharges(hours)));
}
Let me know if the original question is required. I'm using Visual studio 2017 RC.
You can use something like this:
#include <iostream>
#include <windows.h>
//This will set the position of the cursor
void gotoXY(int x, int y) {
//Initialize the coordinates
COORD coord = {x, y};
//Set the position
SetConsoleCursorPosition(GetStdHandle(STD_OUTPUT_HANDLE), coord);
return;
}
void getCursorXY(int &x, int&y) {
CONSOLE_SCREEN_BUFFER_INFO csbi;
if(GetConsoleScreenBufferInfo(GetStdHandle(STD_OUTPUT_HANDLE), &csbi)) {
x = csbi.dwCursorPosition.X;
y = csbi.dwCursorPosition.Y;
}
}
I found it here.
As already written in one of the answers this solution is not platform independent.
But i guess there are similar solutions on other platforms and you can easy set the cursor on the position you want.
Example usage in your main:
for (i = 1; i <= 3; i++)
{
printf("\n%d\t", i);
scanf("%f", &hours);
gotoXY( 20, i + 1);
printf("\t%f\n", calculateCharges(hours));
}
Workarounds for scanf can be found here.
scanf_s always generates a new line upon enter and unfortunately other user input capturing platform independent functions I know of (getc & getchar) do so too. Anyway on Windows it could be done using _getch() from conio header.
#include <conio.h>
#include <stdlib.h>
#include <stdio.h>
int getIntFromUser()
{
char readCharacters[10];
int index = 0;
for (int currentChar = _getch(); currentChar != '\r'; currentChar = _getch())
{
if (currentChar == EOF)
{
// Some error that shouldn't occour in your simple homework program
}
if (index > 9)
{
// Another possible error case where you would start to write beyond 'readCharacters' array
}
// We might as well disallow anything but digits, enter & backspace (You don't need anything else, do you?)
if ((currentChar < '0' || currentChar > '9') && currentChar != '\b')
{
continue;
}
else if (currentChar == '\b')
{
if (index > 0)
{
// Delete last character
printf("\b \b");
readCharacters[index] = '\0';
--index;
}
}
else
{
printf("%c", currentChar);
readCharacters[index] = currentChar;
++index;
}
}
if (index == 0)
{
// User pressed enter without having entered a number, let's give him a zero then
return 0;
}
readCharacters[index] = '\0';
int retVal = atoi(readCharacters);
// Worth noting that the value of converted user given string shouldn't be greater than what a signed int can hold
return retVal;
}
int main(int argc, char* argv[])
{
// Unlike scanf_s this will not generate a new line on enter
printf("getIntFromUser() sample (enter a number)\n");
int someValue = getIntFromUser();
printf(" -- This will be printed on the same line. (someValue is %d)\n\n", someValue);
// scanf_s sample
int anotherValue;
printf("scanf_s() sample (Insert a number.)\n");
scanf_s("%d", &anotherValue);
printf("This will be printed on a new line\n\n");
printf("Press any key to exit.");
_getch();
return 0;
}
EDIT
I feel like the above would become less readable if I were to add a comment over every code line. Instead I'm going to paste some blocks of code 1 by 1.
But first about the _getch function: It waits for the user to type something into the console and then returns the user given char as an int. char implicitly converts to int, so you may compare the _getch result to a character as I did many times in getIntFromUser (e.g. if (currentChar == '\b') { ... }).
You should also know about the values a char can hold and what their values are as an int (check out http://en.cppreference.com/w/cpp/language/ascii).
Going by the table the char '0' would be value 48 as an int, which is what _getch would return if the user were to type a 0.
First declare an array/string of 10 elements. Hope you know about them already. In this case the array is basically a chain of 10 elements that are all of type char, which are also referred to as string.
char readCharacters[10];
An indexer for the string is required.
int index = 0;
Below we have the usual for loop that...
1st: creates a variable of type int and assigns the result of _getch to it.
2nd: will determine if the loop shall keep executing. In this case the loop will break when currentChar is not '\r', which is an escape sequence that represents enter as a character.
3rd: will execute stuff inside once and then update currentChar with a new _getch.
for (int currentChar = _getch(); currentChar != '\r'; currentChar = _getch())
Checks if the user input (retrieved via _getch) is smaller than '0' (value 48 as an int) and greater than '9' (value 57 as an int). If either of them is true it will additionally check if the value of currentChar is not '\b' (value 8 as an int), which is the escape sequence for a backslash.
When that additional check evaluated to true as well then the keyword continue is used. Meaning that the rest of the block in the loop is not executed and instead the loop will start at the top again by getting a new user input and evaluating if the loop is to be continued by checking if obtained currentChar was enter.
if ((currentChar < '0' || currentChar > '9') && currentChar != '\b')
{
continue;
}
NOTE: You might want to read the comments on the else statement before you read these.
When the above if statement was false we get to the next if-statement (actually else if) that we see below.
As mentioned above: '\b' is backslash and if this is the user given char as well as string/array index being greater than 0 we move one character backwards in the console by "printing" '\b' and then write an empty character in order to delete what was written at that place previously. That puts us back to the position we were before so we print another backslash. At this point you might wonder why not just go back to the previous line that scanf_s causes, but that won't work. We must also not forget to replace the last string character with a null terminator and then set the index back by 1.
else if (currentChar == '\b')
{
if (index > 0)
{
// Delete last character
printf("\b \b");
readCharacters[index] = '\0';
--index;
}
}
When we hit this point we know that currentChar is something between 48 and 57 ('0' and '9').
_getch told the program what the user's input was, but we cannot see it in the console unless we print it there. So let's do that.
Also append the user's given character to the string as well as incrementing the index by 1.
else
{
printf("%c", currentChar);
readCharacters[index] = currentChar;
++index;
}
Lastly we call the atoi function that will convert our string/array to an integer.
int retVal = atoi(readCharacters);
I am a newbie to C and I was looking over some questions where I pondered upon a question where we need to scan in values using the users input. Example
1 2 3 45 6 7. So Automatically we scan these values into a 2D array.
One thing that troubles me is what If the user inputs
1 2 3 2 3 Josh, how can we ignore Josh and only scan in the values into the array.
I looked at using getchar and use a flag variable but I am unable to figure out the conundrum of differentiating between the integer and character.
/* This is something that I tried */
#include <stdio.h>
int main(int argc, char *argv[]) {
int a;
int b;
int A[10];
while (((a = getchar()) != '\n') && (b = 0)) {
if (!(a >= "A" && a <= "Z")) {
scanf("%d", A[b]);
}
b++;
}
}
}
I think one good method for achieving what you want is using scanf with the format "%s", which will read everything as a string, effectively splitting the input according to white spaces. From the manual:
s
Matches a sequence of non-white-space characters; the next
pointer must be a pointer to character array that is long
enough to hold the input sequence and the terminating null
byte ('\0'), which is added automatically. The input string
stops at white space or at the maximum field width, whichever
occurs first.
To convert the string to integer, you can use atoi. From the manual:
The atoi() function converts the initial portion of the string
pointed to by nptr to int.
So, if it converts the initial portion of the string into an integer, we can use that to identify what is a number and what's not.
You can build a simple "word detector" for atoi.
Using the function isalpha from ctype.h you can do:
int isword(char *buffer)
{
return isalpha(*buffer);
}
And rewriting your reading program you have:
#include <stdio.h>
#include <ctype.h>
int isword(char *buffer)
{
return isalpha(*buffer);
}
int main(void)
{
char input[200];
int num;
while (1) {
scanf("%s", input);
if (!strcmp(input, "exit")) break;
if (isword(input)) continue;
num = atoi(input);
printf("Got number: %d\n", num);
}
return 0;
}
You should keep in mind that the name isword is fallacious. This function will not detect if buffer is, in fact, a word. It only tests the first character and if that is a character it returns true. The reason for this is the way our base function itoa works. It will return zero if the first character of the buffer is not a number - and that's not what you want. So, if you have other needs, you can use this function as a base.
That's also the reason I wrote a separate function and not:
if (!isalpha(input[0]))
num = itoa(input);
else
continue;
The output (with your input):
$ ./draft
1 2 3 2 3 Josh
Got number: 1
Got number: 2
Got number: 3
Got number: 2
Got number: 3
exit
$
About assigments and &&
while (((a = getchar()) != '\n') && (b = 0))
As I said in a comment, this loop will never work because you're making a logical conjunction(AND) with an assignment that will always return zero. That means the loop condition will always evaluate to false.
In C, assignments return the value assigned. So, if you do
int a = (b = 10);
a will have now hold the value 10. In the same way, when you do
something && (b = 0)
You're effectively doing
something && 0
Which will always evaluate to false (if you remember the AND truth table):
p q p && q
---------------
0 0 0
0 1 0
1 0 0
1 1 1
Your code is completely wrong. I suggest to delete it.
You could use scanf with %d to read in numbers. If it returns 0, there is some invalid input. So, scan and discard a %s and repeat this process:
int num = -1;
while(num != 0)
{
printf("Enter a number, enter 0 to exit:");
if(scanf("%d", &num) == 0) /* If scanf failed */
{
printf("Invalid input found!");
scanf("%*s"); /* Get rid of the invalid input (a word) */
}
}
I want to have a user enter numbers separated by a space and then store each value as an element of an array. Currently I have:
while ((c = getchar()) != '\n')
{
if (c != ' ')
arr[i++] = c - '0';
}
but, of course, this stores one digit per element.
If the user was to type:
10 567 92 3
I was wanting the value 10 to be stored in arr[0], and then 567 in arr[1] etc.
Should I be using scanf instead somehow?
There are several approaches, depending on how robust you want the code to be.
The most straightforward is to use scanf with the %d conversion specifier:
while (scanf("%d", &a[i++]) == 1)
/* empty loop */ ;
The %d conversion specifier tells scanf to skip over any leading whitespace and read up to the next non-digit character. The return value is the number of successful conversions and assignments. Since we're reading a single integer value, the return value should be 1 on success.
As written, this has a number of pitfalls. First, suppose your user enters more numbers than your array is sized to hold; if you're lucky you'll get an access violation immediately. If you're not, you'll wind up clobbering something important that will cause problems later (buffer overflows are a common malware exploit).
So you at least want to add code to make sure you don't go past the end of your array:
while (i < ARRAY_SIZE && scanf("%d", &a[i++]) == 1)
/* empty loop */;
Good so far. But now suppose your user fatfingers a non-numeric character in their input, like 12 3r5 67. As written, the loop will assign 12 to a[0], 3 to a[1], then it will see the r in the input stream, return 0 and exit without saving anything to a[2]. Here's where a subtle bug creeps in -- even though nothing gets assigned to a[2], the expression i++ still gets evaluated, so you'll think you assigned something to a[2] even though it contains a garbage value. So you might want to hold off on incrementing i until you know you had a successful read:
while (i < ARRAY_SIZE && scanf("%d", &a[i]) == 1)
i++;
Ideally, you'd like to reject 3r5 altogether. We can read the character immediately following the number and make sure it's whitespace; if it's not, we reject the input:
#include <ctype.h>
...
int tmp;
char follow;
int count;
...
while (i < ARRAY_SIZE && (count = scanf("%d%c", &tmp, &follow)) > 0)
{
if (count == 2 && isspace(follow) || count == 1)
{
a[i++] = tmp;
}
else
{
printf ("Bad character detected: %c\n", follow);
break;
}
}
If we get two successful conversions, we make sure follow is a whitespace character - if it isn't, we print an error and exit the loop. If we get 1 successful conversion, that means there were no characters following the input number (meaning we hit EOF after the numeric input).
Alternately, we can read each input value as text and use strtol to do the conversion, which also allows you to catch the same kind of problem (my preferred method):
#include <ctype.h>
#include <stdlib.h>
...
char buf[INT_DIGITS + 3]; // account for sign character, newline, and 0 terminator
...
while(i < ARRAY_SIZE && fgets(buf, sizeof buf, stdin) != NULL)
{
char *follow; // note that follow is a pointer to char in this case
int val = (int) strtol(buf, &follow, 10);
if (isspace(*follow) || *follow == 0)
{
a[i++] = val;
}
else
{
printf("%s is not a valid integer string; exiting...\n", buf);
break;
}
}
BUT WAIT THERE'S MORE!
Suppose your user is one of those twisted QA types who likes to throw obnoxious input at your code "just to see what happens" and enters a number like 123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890 which is obviously too large to fit into any of the standard integer types. Believe it or not, scanf("%d", &val) will not yak on this, and will wind up storing something to val, but again it's an input you'd probably like to reject outright.
If you only allow one value per line, this becomes relatively easy to guard against; fgets will store a newline character in the target buffer if there's room, so if we don't see a newline character in the input buffer then the user typed something that's longer than we're prepared to handle:
#include <string.h>
...
while (i < ARRAY_SIZE && fgets(buf, sizeof buf, stdin) != NULL)
{
char *newline = strchr(buf, '\n');
if (!newline)
{
printf("Input value too long\n");
/**
* Read until we see a newline or EOF to clear out the input stream
*/
while (!newline && fgets(buf, sizeof buf, stdin) != NULL)
newline = strchr(buf, '\n');
break;
}
...
}
If you want to allow multiple values per line such as '10 20 30', then this gets a bit harder. We could go back to reading individual characters from the input, and doing a sanity check on each (warning, untested):
...
while (i < ARRAY_SIZE)
{
size_t j = 0;
int c;
while (j < sizeof buf - 1 && (c = getchar()) != EOF) && isdigit(c))
buf[j++] = c;
buf[j] = 0;
if (isdigit(c))
{
printf("Input too long to handle\n");
while ((c = getchar()) != EOF && c != '\n') // clear out input stream
/* empty loop */ ;
break;
}
else if (!isspace(c))
{
if (isgraph(c)
printf("Non-digit character %c seen in numeric input\n", c);
else
printf("Non-digit character %o seen in numeric input\n", c);
while ((c = getchar()) != EOF && c != '\n') // clear out input stream
/* empty loop */ ;
break;
}
else
a[i++] = (int) strtol(buffer, NULL, 10); // no need for follow pointer,
// since we've already checked
// for non-digit characters.
}
Welcome to the wonderfully whacked-up world of interactive input in C.
Small change to your code: only increment i when you read the space:
while ((c = getchar()) != '\n')
{
if (c != ' ')
arr[i] = arr[i] * 10 + c - '0';
else
i++;
}
Of course, it's better to use scanf:
while (scanf("%d", &a[i++]) == 1);
providing that you have enough space in the array. Also, be careful that the while above ends with ;, everything is done inside the loop condition.
As a matter of fact, every return value should be checked.
scanf returns the number of items successfully scanned.
Give this code a try:
#include <stdio.h>
int main()
{
int arr[500];
int i = 0;
int sc = 0; //scanned items
int n = 3; // no of integers to be scanned from the single line in stdin
while( sc<n )
{
sc += scanf("%d",&arr[i++]);
}
}