I'm trying to reference global variables in a function called "userval", and then modify those variables based on user input. Do I need to return these variables at the end of my function?
I am attempting to check my code by printing these global variables within the main function. However, I keep getting random characters.
Here is my code below.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
// Global Constants
#define MAX_PSWD_LEN 80
char password = 0;
int len = 0;
int upper = 0;
int lower = 0;
int digit = 0;
int special = 0;
int main(int argc, char *argv[]) {
userval();
printf(&len);
printf(&upper);
printf(&lower);
printf(&digit);
printf(&special);
}
int userval() {
printf("Please enter the minimum length of your password: \n");
scanf("%d", &len);
printf("Please enter the minimum number of uppercase letters in your password: \n");
scanf("%d", &upper);
printf("Please enter the minimum number of lowercase letters in your password: \n");
scanf("%d", &lower);
printf("Please enter the minimum number of decimal digit characters in your password: \n");
scanf("%d", &digit);
printf("Please enter the minimum number of special characters in your password: \n");
scanf("%d", &special);
printf("Thank you. \n");
return len, upper, lower, digit, special;
}
That's not how the printf function works.
The first parameter is a format string. It contains any static text you want to print along with format specifiers for any values you want to print.
For example, if you want to print only a integer followed by a newline, the format string you would use is "%d\n", where %d is the format specifier for an integer and \n is a newline character.
Any subsequent parameters are used to fill in the format. In the case of the values you want to print, you would do the following:
printf("%d\n", len);
printf("%d\n", upper);
printf("%d\n", lower);
printf("%d\n", digit);
printf("%d\n", special);
The correct way to use it would be to
printf("%p\n",(void*)&len);
But this would print the address of the variable - most likely you want to print the value of the variable. (Holds for other int` variables also in your example).
printf("%d\n",len);
While using printf the first argument is the format string and the rest of the arguments are 0 or more variables (as dictated by format string).
From standard
int printf(const char * restrict format, ...);
This is the signature of the printf function. Also regarding the format specifier §7.21.6.1
The format shall be a multibyte character sequence, beginning and ending in its initial shift state. The format is composed of zero or more directives: ordinary multibyte characters (not %), which are copied unchanged to the output stream; and conversion specifications, each of which results in fetching zero or more subsequent arguments, converting them, if applicable, according to the corresponding conversion specifier, and then writing the result to the output stream.
Instead of any output stream it is stdout for printf.
Related
I'm a student and was given this function by our teacher to get user input. It's used for integers but I'm confused why there's a char data type. If someone could explain this function that would be great.
#pragma warning (disable: 4996)
int getNum(void)
{
char record[121] = {0};
int number = 0;
fgets(record, 121, stdin);
if( sscanf(record, "%d", &number) != 1)
{
number = -1;
}
return number;
}
When calling we do this for example:
int age = 0; //we always have to initialize
age = getNum(); //this is how we call the function to get user input
The char record[121] variable is actually an array of characters, otherwise known as a string. It is used here so that, if an erroneous input is given (instead of a valid integer input), then the input stream is still cleared by the fgets() call, which takes in all characters up to (and including) the newline character.
The sscanf function then attempts to extract a valid integer from the read character string and, if it fails, then an error value (-1) is assigned to signal that fact. If it succeeds, it assigns the given input value to number, in the same way that a successful call to scanf("%d", &number) would.
However, just using such a simple scanf() call would potentially leave 'unprocessed' characters in the input stream, which could cause problems for any subsequent input operations (or require the input stream to be cleared).
Using such a technique will also prevent undesired/unexpected effects caused by a user typing extra characters after a valid input. For example, the following (deliberately bad) code will cause a 'surprise' if you enter 1a in response to the first prompt:
#include <stdio.h>
int main()
{
int number = 0, number2 = 0;
printf("Enter first number: ");
scanf("%d", &number);
printf("Enter second number: ");
scanf("%d", &number2);
printf("Given numbers were: %d and %d.\n", number, number2);
return 0;
}
However, using the getNum() function your teacher provided, the following code is much more robust:
int main()
{
int number = 0, number2 = 0;
printf("Enter first number: ");
number = getNum();
printf("Enter second number: ");
number2 = getNum();
printf("Given numbers were: %d and %d.\n", number, number2);
return 0;
}
The program is reading lines from a file with fgets, which uses a string as storage buffer for the line. After that it processes the line to find a string that can be read as an int (%d), checks if that operation went OK, and returns the integer if so.
The function is reading input as text, storing it to the character array record. It then converts that string into an equivalent integer value using sscanf.
This is a bit safer than just using scanf directly - it ensures that if the user types non-numeric input, it's still consumed from the input stream so that it doesn't cause problems later on.
Only my first line of input request user to key in the value. Input B not request user to key in and shows wrong total.
#include <stdio.h>
#include <stdlib.h>
//BASIC CALCULATION INPUT 2 INTEGER ONE BY ONE
int main(int argc, char *argv[])
{
int a,b,c;
//REQUEST ONE INPUT
printf("Integer A: \n");
scanf("%a",&a);
//REQUEST ONE INPUT
printf("Integer B: \n");
scanf("%b",&b);
c=a+b;
//DISPLAY AMOUNT INTEGER
printf("Total: &c",c);
system("PAUSE");
return 0;
}
Your both scanf statements are wrong!
It should be
scanf ("%d",&a);
scanf ("%d",&b);
For taking user input a and b of type integer use %d.
&a is the reference (address) of identifer a which holds the value of a.
And also user output printf statement for integer c should be
printf("%d",c);
Add a space in the scanf to discard all whitespace before matching an integer. Such as a
scanf(" %b",&b);
^^^
space in the scanf
There is no format specifier like %b that's why Input B doesn't request user to key in.I think this this will help you.
#include <stdio.h>
#include <stdlib.h>
//BASIC CALCULATION INPUT 2 INTEGER ONE BY ONE
int main(int argc, char *argv[])
{
int a,b,c;
//REQUEST ONE INPUT
printf("Integer A: \n");
scanf("%d",&a);
//REQUEST ONE INPUT
printf("Integer B: \n");
scanf("%d",&b);
c=a+b;
//DISPLAY AMOUNT INTEGER
printf("Total: &c",c);
system("PAUSE");
return 0;
}
Both of your scanf statements use wrong format specifiers. Thus undefined behaviour.
scanf("%a",&a);
and
scanf("%b",&b);
a expects a float*as its argument, but you are passingint*. There's no format specifierb` in standard C either.
Use %d to scan int's.
Better yet, avoid scanf altogether and use fgets and parse the line instead.
Another problem is your printf statement:
printf("Total: &c",c);
is wrong too. It should be:
printf("Total: %c",c);
to print the value of c.
I'm new to C and have been playing about and having created a basic input output program for integers (See below) I was wondering how I would go about making my program omit characters from an inputted number and then print the number without said omitted characters so for example if I wanted it to omit non-primes 1234567 entered would print 2357. Here is my current code:
#include <stdio.h>
int main() {
int number;
// printf() dislpays the formatted output
printf("Enter an integer: ");
// scanf() reads the formatted input and stores them
scanf("%d", &number);
// printf() displays the formatted output
printf("You entered: %d", number);
return 0;
}
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. :)
Program:
#include "stdafx.h"
#include "stdio.h"
#include "string.h"
//#include "iostream"
const int IDLEN = 10;
const int POLARITYLEN = 3;
const int MAXSTOCKITEMS = 10;
struct TransistorRec {
char manufacturersID[IDLEN + 1];
char polarity[POLARITYLEN + 1];
float power;
float gain;
int stock;
};
typedef struct TransistorRec Transistor;
struct StockRec{
int size;
Transistor stocklist[MAXSTOCKITEMS];
};
typedef struct StockRec Stock;
int main()
{
int total, i;
struct TransistorRec a[10];
char x, y;
printf("How many transistors: ");
scanf_s("%i", &total);
if (total >MAXSTOCKITEMS){
printf("too much!! repeat");
scanf_s("%i\n", &total);
}
for (i = 0; i < total; i++)
{
printf("Enter manufacturer's ID of transistor:");
scanf_s("%s",a[i].manufacturersID);
printf("Enter polarity of transistor: ");
scanf_s("%s",a[i].polarity);
printf("Enter power of transistor: ");
scanf_s("%f",a[i].power);
printf("Enter gain of transistor: ");
scanf_s("%f",a[i].gain);
printf("Enter current stock of transistor: ");
scanf_s("%i",a[i].stock);
}
return 0;
}
My assignment:
Write a C function that reads from the user (keyboard) the information about the transistors in stock,
and stores this information in the Stock structure. A sample set of input data (manufacturer’s ID,
polarity, maximum power, current gain, number in stock) is:
2N2222 NPN 0.5 75 23
BC559 PNP 0.5 125 7
TIP31B NPN 40.0 20 11
Problem:
When i input my values(strings) through scanf_s to a structure , the program will stop.
Quoting from the documentation of scanf_s,
Remarks:
[...]
Unlike scanf and wscanf, scanf_s and wscanf_s require the buffer size to be specified for all input parameters of type c, C, s, S, or string control sets that are enclosed in []. The buffer size in characters is passed as an additional parameter immediately following the pointer to the buffer or variable.
So, the scanf_ss:
scanf_s("%s",a[i].manufacturersID);
scanf_s("%s",a[i].polarity);
need a third argument specifing the buffer size as mentioned in the above quote. So, use
scanf_s("%s",a[i].manufacturersID,sizeof(a[i].manufacturersID));
scanf_s("%s",a[i].polarity,sizeof(a[i].polarity));
or
scanf_s("%s",a[i].manufacturersID,_countof(a[i].manufacturersID));
scanf_s("%s",a[i].polarity,_countof(a[i].polarity));
There are other errors too. These:
scanf_s("%f",a[i].power);
scanf_s("%f",a[i].gain);
scanf_s("%i",a[i].stock);
needs to be
scanf_s("%f",&a[i].power);
scanf_s("%f",&a[i].gain);
scanf_s("%i",&a[i].stock);
as the %f and %i expects a float* and int* respectively, and not a float and int.
As for why the first two scanf_s did not require a &, because ,in C, array names "decay" to a pointer to its first element. a[i].manufacturersID and a[i].polarity are arrays and the "decay" automatically happens.
You should also remove the \n from
scanf_s("%i\n", &total);
because \n is considered as a whitespace character and this character will discard any number of whitespace characters including none until the first non-whitespace character as specified in the C11 standard:
7.21.6.2 The fscanf function
[...]
A directive composed of white-space character(s) is executed by reading input up to the first non-white-space character (which remains unread), or until no more characters can be read. The directive never fails.