The practical difference of %c vs %s - c

I'm required to do a seafood menu as an assignment in my college. I will provide the relevant code here:
int main ()
{
int a,d;
float c ;
char b,s,S,m,M,l,L;
printf ("\n\t\t\t\tSeafood Menu\n");
printf ("----------------------------------------------------------------------------------\n");
printf ("\t\t\t\t\t\t Dish Size\n");
printf ("Item Number\t Seafood Dish\t Small\t Medium Large\t\n");
printf ("----------------------------------------------------------------------------------\n");
printf ("\t 1\t Fried \t20.00 \t\t 40.00 \t\t 55.00\n");
//Continue with a bunch of menu here
printf (" Enter item number :");
scanf ("%d",&a);
printf (" Enter dish size (S/M/L) :");
scanf ("%s",&b);
if ((a==1)&&((b=='s')||(b=='S')))
{c=20.00;}
//continue with a bunch of condition to choose the price per dish stated in the menu
printf (" Enter dish quantity :");
scanf ("%d",&d);
printf (" Price per dish size :RM%.2f\n\n\n",c);
return 0;
}
When I tried to change the format identifier in this to %c, it just stopped accepting input for that particular scanf.
printf ("Enter dish size (S/M/L):");
scanf ("%s",b);
I would like to attach images but it seems that I am not allowed to do so 'll leave two links:
Normal, using %s: and abnormal, using %c
I'm curious about why it doesn't work when I use %c while %s works? As all I enter there is just character. Please enlighten me.

I'm curious about why it doesn't work when I use %c while %s works?
Well, guess what, it did not work, even if it appeared to.
As mentioned in the man page, for scanf(), the type of arguments are
%c expects a pointer to char
%s expects a pointer to the initial element of an char array.
Quoting C11, chapter §7.21.6.2, (emphasis mine)
c Matches a sequence of characters of exactly the number specified by the field
width (1 if no field width is present in the directive).
If no l length modifier is present, the corresponding argument shall be a
pointer to the initial element of a character array large enough to accept the
sequence. No null character is added.
s Matches a sequence of non-white-space characters.
If no l length modifier is present,the corresponding argument shall be a
pointer to the initial element of a character array large enough to accept the
sequence and a terminating null character, which will be added automatically.
In your case,
providing %s is wrong, as a single char is one element too short to scan and hold an array, that is null-terminated. This causes undefined behavior as memory overrun happens.
%c should have "worked" , if the input stream did not have the newline stored from previous ENTER key press. If you clean off the input stream of all pending inputs, you'll see it works. Use something like
scanf(" %c", &b); //the whitespace "eats up" all the previous whitespaces
You can see this previous answer of mine, for details on the same.

Related

Why is scanf failing to read inputs correctly?

I cant figure out whats wrong. Am i using format specifiers in wrong way? Someone please help i am very new to coding.
#include <stdio.h>
#include <stdlib.h>
int main()
{
char name[20];int age;char grade;double gpa;char area[10];
printf("User Input\n");
printf("Enter your name: ");
fgets(name,20,stdin);
printf("Your name is %s\n",name);
printf("Enter your age: ");
scanf("%d\n",&age);
printf("Your age is %d\n",age);
printf("Enter you grade: ");
scanf("%c\n",&grade);
printf("Your grade is %c\n",grade);//Why is this giving an int output?
printf("Enter your gpa: ");
scanf("%f\n",&gpa);
printf("Your gpa is %f\n",gpa);
printf("Enter your area: ");
scanf("%s\n",&area);
printf("Your area is %s",area);//This shows grade input
return 0;
}
Output
You use fgets correctly when reading name. I'd recommend also using fgets for all your other inputs, and then parsing the intended values out of them. For example:
char age_str[20];
fgets(age_str, 20, stdin);
age = strtol(age_str, NULL, 10);
This is preferable to using scanf directly for non-string inputs since if input fails to match a format string, it will remain in stdin and screw up the other scanf calls.
If you would like to use scanf correctly:
Check its return value to see if it matches the number of format specifiers in the string. If not, some inputs were not successfully read. You may want to use a do/while loop for this.
Begin your format strings with a space, as in " %c", so that any whitespace remaining in stdin will be skipped over.
Don't end your format strings with a newline.
Some things to remember about scanf:
Most conversion specifiers like %s, %d, and %f will skip over leading whitespace - %c and %[ will not. If you want to read the next single non-whitespace character, use " %c" - the leading blank tells scanf skip over any leading whitespace before reading the next non-whitespace character;
For what you are trying to do, you should not use \n in your format strings - it will cause scanf to block until you enter a non-whitespace character;
You do not need to use the & operator on array expressions like area; under most circumstances, array expressions are converted to pointer expressions1. Honestly, you should read area the same way you read name, using fgets (and you should always check the result of fgets), or you should specify the maximum field width in the specifier: scanf( "%9s", area ); (a 10-element array can hold up to a 9-character string, since one element has to be reserved for the string terminator);
You should get in the habit of checking the result of scanf - it will return the number of successful conversions and assignments. For example, scanf( "%d %d", &x, &y ) will return 2 if both x and y are read successfully. It will return EOF if end-of-file is signaled or there's a read error.
scanf will read up to the next character that doesn't match the conversion specifier - IOW, if you're using %d, then scanf will skip over any leading whitespace, then read up to the next character that isn't a decimal digit. That character is left in the input stream. This means if you're using %d and type in 123e456, scanf will read up to that 'e' character and assign 123 to the target. If you try to read again with %d, scanf will immediately stop reading on that e and return 0 without assigning anything to the target (this is called a matching failure). This will continue until you remove that 'e' from the input stream (such as with getchar or fgetc or scanf with the %c specifier, etc.
You need to make sure the types of the arguments match the format specifier. %s expects an argument of type char *, %d expects int *, %f expects float *. %x expects unsigned int *, %lf expects double *, etc.
This is one of the "deeply unintuitive" aspects of C I was talking about in my comment.

What is the specific reason for the runtime error I'm getting here?

#include<stdio.h>
#include<string.h>
void main()
{
char a,b,c;
printf("Enter alien names:\n");
scanf("%s\n%s\n%s\n",a,b,c);
printf("The alien names are %s, %s and %s. A meteor hit %s's spaceship. A star scratched %s\'s spaceship. But %s fixed %s and %s\'s spaceships. The three became friends and are from the planet BYG (which means BLUE YELLOW GREEN)",a,b,c,a,b,c,a,b);
}
What is the specific reason for the runtime error I'm getting here?
To solve this issue you should simply consider to use strings (arrays of chars) to contain the different names.
Here is an example how to do that:
void main()
{
// The string "a" can contain up to 100 symbols (chars).
char a[100];
printf("Enter an alien name:\n");
scanf("%s",a);
printf("The alien name is %s.", a);
}
The difference between "char a" and "char a[100]" is that in the first case the variable "a" corresponds to a single character and in the second it corresponds to a string - an array of chars which can contain up to 100 characters.
The posted code has undefined behavior, because the variables a, b, and c are of type char, while the %s conversion specifier in the call to scanf() is expecting a pointer to the first element of a character array that can hold the input string. Mismatched conversion specifers and arguments in a scanf() call lead to undefined behavior, and attempting to write too many characters into the receiving array causes undefined behavior.
The first problem can be fixed by declaring a, b, and c as arrays large enough to hold expected input:
char a[100], b[100], c[100];
...
scanf("%s\n%s\n%s\n", a, b, c);
Note that arrays decay to pointers to their first elements in most expressions, including function calls, so here a is a pointer to the first element of the character array a[]; this is equivalent to &a[0].
There is still a possibility for undefined behavior if the user enters too many characters. To avoid this, always specify a maximum width when using scanf() to read user input into a string. Note here that the specified width is the maximum number of characters that will be read for that input item, not including the null terminator, \0, which will be automatically added by scanf(), so the maximum width must be at least one less than the size of the receiving array:
scanf("%99s\n%99s\n%99s\n", a, b, c);
But if you compile and run this code, you will find that it does not behave as expected. After the third name is entered, the program will continue waiting for more input. This is because the \n character is a whitespace character, and when scanf() encounters a whitespace character in a format string, it reads and discards zero or more whitespace characters in the input until a nonwhitespace character is encountered, or until no more characters can be read. The %s directive tells scanf() to read characters until a whitespace character is encountered. So when the user presses Enter after the final name, scanf() completes matching input characters for the final name and returns the \n character to the input stream; then the \n is reached in the above format string, and scanf() matches the aforementioned \n character in the input stream, and any further whitespace characters that are encountered. This will end if the user enters another nonwhitespace character, or signals end-of-file from the keyboard (e.g., with Ctrl-D or Ctrl-Z).
To avoid this complication, remember that it is almost never correct to end a scanf() format string with a whitespace character. Also, there is no need to use \n rather than a space character, since both are simply interpreted as whitespace directives by scanf():
scanf("%99s %99s %99s", a, b, c);
It would further improve the posted code if the return value from the call to scanf() were checked before attempting to use the input. Since scanf() returns the number of successful assignments made, this value should be 3:
#include <stdio.h>
#include <string.h>
int main(void)
{
char a[100], b[100], c[100];
printf("Enter alien names:\n");
int ret_val = scanf("%99s %99s %99s", a, b, c);
if (ret_val == 3) {
printf("The alien names are %s, %s and %s. A meteor hit %s's "
"spaceship. A star scratched %s\'s spaceship. But %s "
"fixed %s and %s\'s spaceships. The three became friends "
"and are from the planet BYG (which means BLUE YELLOW GREEN)\n",
a, b, c, a, b, c, a, b);
} else {
puts("Input error");
}
}
What is the specific reason for the runtime error I'm getting here?
The function scanf using the format specifier %s expects to be passed the address of a char array, in which to place the input data. For example an array such as
char a[100];
However, you pass simple char variables a and b and c which can hold values in the range -128 to 127, or 0 to 255, depending on whether the implementation's char is signed or unsigned.
These variables were not even initialised, so indeterminate values were passed to scanf. But even if they had been initialised, it is very likely that the values passed will cause a segfault, when used as addresses.
My compiler issued 2 warnings for each of a, b and c passed to scanf.
warning C4477: 'scanf' : format string '%s' requires an argument of type 'char *', but variadic argument 1 has type 'int'
warning C4700: uninitialized local variable 'a' used
Please enable and act on all compiler warnings.
//There are things that shoudn't be there. Im not a pro but this is what I think.
#include<stdio.h>
#include<string.h>//you have include this library but you didn't use a function from it.
//I think what you want to do is use the str functions like strcpy
//but in this case you don't need to use it.
void main()
{
char a[25],b[25],c[25];//Here you declared a character a, b and c. But if you want to store a string, you have to declare an array of characters. So instead of a, b, c, it's a[someValue], b[someValue] and c[someValue].
//Declare an array with a size that you think will cover the whole "alien name". e.g. a[25]..
//but i don't know, maybe you did it on purpose. Maybe you just want to name the aliens with one character like A, B, C. But if you want to name the aliens with a long name, you must declare an array.
printf("Enter alien names:\n");
scanf("%s\n%s\n%s\n",a,b,c);//You don't need to put the "\n" between those "%s". "\n" means "newline". It will work without it because scanf automatically reads next set of characters when it meets white space of newline.
//--so you can remove "\n" in there and replace it with space. But you can leave it there also but you really have to remove the last "\n" because scanf will search again for the next
//--new line before it will end asking for input and pressing enter will not work because you have to type another set of characters before scanf will read the last "\n" that you put at scanF.
//Another mistake here is the format specifier that you used (%s). It doesn't match declaration because you declare char a, b, c, that will only store one character each.
//In case that you're really just storing one character each alien's name, you have to use the "%c" instead of "%s" and you must pass the reference of the char variable in
//--scanf, e.g. scanf("%c %c %c", &a, &b, &c);
//Just remember that if you plan on storing a string or a long name there, you must declare an array like I said at the beginning.
//--and if it's an array, you don't need to include the '&' on every variable when you're passing it in scanF.
//There's nothing wrong here if you're alien's names are string.
printf("The alien names are %s, %s and %s. A meteor hit %s's spaceship. A star scratched %s\'s spaceship. But %s fixed %s and %s\'s spaceships. The three became friends and are from the planet BYG (which means BLUE YELLOW GREEN)",a,b,c,a,b,c,a,b);
}
The specific reason for the runtime error is this line:
scanf("%s\n%s\n%s\n",a,b,c);
The %s conversion specifier tells scanf to read a sequence of non-whitespace characters from the input stream (skipping over leading whitespace) and store that sequence to an array of char pointed to by the corresponding argument. The problem is that a, b, and c are not pointers to char; they're single char objects that haven't been initialized. The odds of any of them containing a value that corresponds to an address that scanf can write to is almost non-existant.
First, change the declarations of a, b, and c tochar a[SOME_LENGTH] = {0}; // initialize array contents to 01
char b[SOME_LENGTH] = {0};
char c[SOME_LENGTH] = {0};
where SOME_LENGTH is a number that's long enough to contain the longest string you expect to enter plus one extra space for the string terminator. IOW, if the longest string you intend to read is 10 characters long, then your declarations need to be
char a[11] = {0};
char b[11] = {0};
char c[11] = {0};
Secondly, change your scanf call to
scanf( "%(SOME_LENGTH-1)s %(SOME_LENGTH-1)s %(SOME_LENGTH-1)s", a, b, c );
where (SOME_LENGTH-1) is the length of your buffer minus 1. Again, assuming SOME_LENGTH is 11:
scanf( "%10s %10s %10s", a, b, c );
This will help prevent a buffer overrun in the event you enter a string longer than what the buffer is sized to hold.
Both the %s conversion specifier and a blank space in the format string tell scanf to consume and discard any leading whitespace. You can run into trouble specifying whitespace characters in the format string.
Additional notes:
main returns int, not void - change your main to
int main (void)
{
...
}
If there are fewer elements in the initializer than there are in the array, then excess elements are initialized to 0. So in this case, the first element is *explicitly* initialized to 0, and the remaining elements are *implicitly* initialized to 0.

problems with scanf and conversion specifiers

Here you can see my source code:
#include <stdio.h>
int main()
{
char yourname;
char yoursex;
int yourage = 0;
printf("Hey, what's your name?\n");
printf("My name is: ");
scanf("%s", &yourname);
printf("Oh, hello %s! \n\n", &yourname);
printf("Are you a boy or a girl?: ");
scanf("%s", &yoursex);
printf("Nice to know you are a %s! \n\n", &yoursex);
printf("How old are you %s? I am ", &yourname);
scanf("%d", &yourage);
printf("I see you are %d, you have many years then!", &yourage);
return 0;
}
I was trying things that I didn't knew, and strangely it is not working for me. What's the problem? Also, why it needs to be %s and not %c? If I use %c instead it does not work!
Where it says:
How old are you %s? instead of putting my name, it says ''oy''
and instead of showing my age in the last line, it shows a big number.
These are the very basics of C Programming, and I strongly advise you to get a decent book - The C Programming Language by Dennis Ritchie would be a good start.
There are numerous errors in your code.
A char can contain only one character, like 'A', or 'a' or something like that. When you're scanning a name, it is going to be a group of characters, like 'E', 'd', 'd', 'y'. To store multiple characters, you need to use a character array. Also, the format specifier used to scan/print characters is %c, %s is for when you need to scan a group of characters, also called a string into an array.
When you use printf, you do not supply a pointer to the variable you are trying to print (&x is a pointer to variable x). The pointer is a 32/64-bit integer, which is likely why you see a random integer when trying to print. printf("%c\n", charVar) is sufficient.
scanf does not need an & while using %s as the format specifier, assuming you have passed a character array as the argument. The reason is, scanf needs to know where to store the data you are reading from the input - and that is given by a pointer to the memory location. When you need to scan an integer, you need to pass an &x - which means, pointer to memory location of x. But when you pass a character array, it is already in the form of a memory address, and doesn't need to be preceded by an ampersand.
I once again recommend you look up some decent tutorials online, or get a book (the one I mentioned above is a classic). Type the examples as given in the material. Experiment. Have fun. :)
%s is for reading a string -- multiple characters delimited by whitespace. %c is for reading a single char.
You declare your yourname and yoursex vars as characters, and then try to read strings into them. The string read will overwrite random other things in the stack frame and misbehave or crash.
You want to declare yourname and yoursex as character arrays, so they can hold strings:
char yourname[32];
char yoursex[32];
then, when reading into them, you want to include a length limit so they don't overflow:
scanf("%31s", yourname);
This is a single character:
char yourname;
But %s indicates that the variable is a string (i.e., an array of characters terminated by a NUL). That's why you need %c. If you really did mean to use a string, then define the variable like
char yourname[32]; /* just pick a big enough size */
Also, you are correct to use the address of the variable with scanf(), but printf() needs the value. So instead of
printf("I see you are %d, you have many years then!", &yourage);
use
printf("I see you are %d, you have many years then!", yourage);
The "big number" is the memory address.
Make sure you read the comments in code!
#include <stdio.h>
int main()
{
char yourname[10];
char yoursex[5]; // boy or girl + null terminator
int yourage = 0;
printf("Hey, what's your name?\n");
printf("My name is: ");
scanf("%s", &(*yourname)); // & and * cancel each other out,
// thus take a look at the next scanf()
printf("Oh, hello %s! \n\n", yourname); // yourname is now an array
printf("Are you a boy or a girl?: ");
scanf("%s", yoursex);
printf("Nice to know you are a %s! \n\n", yoursex);
printf("How old are you %s? I am ", yourname);
scanf("%d", &yourage); // ok
printf("I see you are %d, you have many years then!", yourage); // here you don't
// need the address of the variable!
return 0;
}
The expression char yourname; only holds space for a single character, so quite likely you end up corrupting the memory space when scanning for yourname. You should allocate a bigger buffer and make sure that you don't overrun its length by setting a maximum number of characters to be read with the scanf function; as described in some of the other answers.
The fact that the following printf print correctly the name doesn't mean that the memory doesn't get corrupted; as C/C++ don't really check the boundary of any strings or arrays used at runtime.
As suggested by others, starting by reading a good book about C and/or C++ wouldn't a bad idea.

having trouble with a "\n" and scanf

Here is the code
printf("\n");
printf("Enter a integer vaule:");
scanf("%d" , &num3);
printf("You entered: %015d", num3);
printf("Enter a float value:");
scanf("%f", &deci3);
printf("You entered: %15.2f", deci3);
printf("\n");
the output is
Enter a integer vaule:4.4
You entered: 000000000000004
Enter a float value:You entered: 0.40
The problem is this code is not stopping at
printf("Enter a float value:");
and this scanf
scanf("%f", &deci3);
seems to be getting its value from the previous scanf
The %d conversion stops wherever the integer stops, which is a decimal point. If you want to discard the input there, do so explicitly… getc in a loop, fgets, or such. This also allows you to validate the input. The program should probably complain about 4.4.
The scanf function works this way per the specification:
An input item shall be defined as the longest sequence of input bytes (up to any specified maximum field width, which may be measured in characters or bytes dependent on the conversion specifier) which is an initial subsequence of a matching sequence. [Emphasis added.]
In your example, the following C string represents the contents of stdin when the first scanf call requests input: "4.4\n".
For this initial call, your format string consists of a single specifier, %d, which represents an integer. That means that the function will read as many bytes as possible from stdin which satisfy the definition of an integer. In your example, that's just 4, leaving stdin to contain ".4\n" (if this is confusing for you, you might want to check out what an integer is).
The second call to scanf does not request any additional input from the user because stdin already contains ".4\n" as shown above. Using the format string %f attempts to read a floating-point number from the current value of stdin. The number it reads is .4 (per the specification, scanf disregards whitespace like \n in most cases).
To fully answer your question, the problem is not that you're misusing scanf, but rather that there's a mismatch between what you're inputting and how you're expecting scanf to behave.
If you want to guarantee that people can't mess up the input like that, I would recommend using strtol and strtod in conjunction with fgets instead.
This works, but it dont complains if you type 4.4 for the int
#include <stdio.h>
int main() {
char buffer[256];
int i;
float f;
printf("enter an integer : ");
fgets(buffer,256,stdin);
sscanf(buffer, "%d", &i);
printf("you entered : %d\n", i);
printf("enter a float : ");
fgets(buffer,256,stdin);
sscanf(buffer, "%f", &f);
printf("you entered : %f\n", f) ;
return 0;
}
use a fflush(stdin) function after the fist scanf(), this will flush the input buffer.

Adding data to arrays in C

I have a few arrays that I wanted to update. The problem is that when data is added to the array, some variables are updated to 0.
my code to add data:
void addStockItem(){
system(CLS_NAME);
printf("New Item\n\n");
printf("New Item Name : ");
fflush(stdin);
scanf(" %[^\n]s",&*itemName[totalItem]);
//itemName[totalItem][30] = strupr(itemName[totalItem]);
fflush(stdin);
printf("New Item Price : ");
scanf("%f",&item_Price_Profit[totalItem][0]);
printf("New Item Profit : ");
scanf("%f",&item_Price_Profit[i][1]);
printf("New Item Qty : ");
scanf("%d",&itemQuantity[totalItem][0]);
itemQuantity[totalItem][1]=0;
itemQuantity[totalItem][2]=0;
++totalItem;
allStocks();
}
the data,
int totalItem=13,itemQuantity[][3]={8,0,0,9,0,0,11,0,0,0,0,0,20,0,0,22,0,\
0,16,0,0,18,0,0,9,0,0,7,0,0,5,0,0,12,0,0,0,0,0},sessionQuantity;
float item_Price_Profit[][2]={1,0.5,2,0.2,3,0.2,4,0.2,5,0.5,6,0.8,7,0.5,8,0.2,9,\
0.2,10,0.2,11,0.5,12,0.8,13,0.9};
char itemName[][30]={"STABILO PENCIL 2B","STABILO PEN 0.5",\
"STABILO ERASER","STABILO RULER","STABILO TEST PAD","STABILO BOOK","STABILO SCISSORS","STABILO SHARPENER","STABILO GLUE","STABILO CHALK","STABILO MARKER PEN","OXFORD DICTIONARY","STABILO HIGHLIGHTER"};
full code: http://pastebin.com/jjuCCrjz
[EDIT]
Everything work as they intended to after I changed itemQuantity[][3] to itemQuantity[100][3], item_Price_Profit[][2] to item_Price_Profit[100][2] and itemName[][30] to itemName[100][30]. What could possibly my mistake other than scanf?
I can't access pastebin from my work computer, so I'll have to go by what's posted.
Several issues off the bat:
fflush is only defined to work on output streams, not input streams; the behavior of fflush(stdin) is undefined (it will do something, but probably not what you want). If you need to clear out garbage from the input stream, you'll need to consume it using getchar() or fgets() or similar until you see a newline or some other indicator that you've cleared out the garbage. The %f and %d conversion specifiers will skip over any leading whitespace, and by having the blank before the %[ conversion specifier will also cause any leading whitespace to be skipped. So ditch the fflush calls altogether.
scanf(" %[^\n]s",&*itemName[totalItem]); - this looks confused. Unless you expect the input to always have a trailing s character, the conversion specifier should simply be %[^\n]. The &* in front of itemName is redundant; you just need to write
scanf(" %[^\n]", itemName[totalItem]);Although you should probably put a field width specifier in there:
scanf(" %30[^\n]", itemName[titalItem]);
to avoid buffer overflow.
You're accessing all your data items (totalItem, itemName, item_Price_Profit, etc.) as global variables. This is usually a recipe for heartburn. Ideally, functions and their callers should not share state via globals; rather, they should communicate through parameters, return values, and exceptions (where supported). Something more like
void addStockItem(char *name, float *price, float *profit, float *quantity)
{
...
scanf(" %30[^\n]", name);
...
scanf("%f", price);
...
scanf("%f", profit);
...
scanf("%d", quantity);
} which would be called like
addStockItem(itemName[totalItem],
&item_Price_Profit[totalItem][0],
&item_Price_Profit[i][1],
&itemQuantity[totalItem][0]);
Your data structures feel really hinky to me, but that's likely because I can't see your entire program.
This:
scanf(" %[^\n]s",&*itemName[totalItem]);
can't be right. This is passing a character converted to a pointer, where scanf() expects a pointer to character. You probably mean:
scanf(" %[^\n]s", itemName[totalItem]);

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