Why can i not use %s instead of %c? - c

The whole function the question is about is about giving a two dimensional array initialized with {0} as output and making a user able to move a 1 over the field with
char wasd;
scanf("%c", &wasd);
(the function to move by changing the value of the variable wasd is not important i think)
now my question is why using
scanf("%s", &wasd);
does only work partly(sometimes the 1 keeps being at a field and appears a 2nd time at the new place though it actually should be deleted)
and
scanf("%.1s", &wasd);
leads to the field being printed out without stop until closing the execution program. I came up with using %.1s after researching the difference between %c and %s here Why does C's printf format string have both %c and %s?? If one can figure out the answer by reading through that, i am not clever or far enough with c learning to get it.
I also found this fscanf() in C - difference between %s and %c but i do not know anything about EOF which one answer says is the cause of the problem so i would prefer getting an answer without it.
Thank you for an answer

Simple as that, %s is the conversion for a (non-empty) string. A string in C always ends with a 0 byte, so any non-empty string needs at least two bytes. If you pass a pointer to a single char variable, scanf() will just overwrite whatever is in memory after that variable -- you cause undefined behavior and anything can happen.
Side note, scanf("%s", ..), even if you give it an array of char, will always overflow the buffer if something longer is entered, therefore causing undefined behavior. You have to include a field width like
char str[10];
scanf("%9s", str);
Best is not to use scanf() at all. For your single character input, you can just use getchar() (be aware it returns an int). You might also want to read my beginners' guide away from scanf.

A char variable can hold only one byte of memory to hold a single character. But a string (array of characters) is different from a char variable as it is always ended with a null character \0 or numeric 0. So in scanf you specifically mentioned whether you are reading a character or a string so that scanf can add a null character at the end of a string. So you are not suppose to use a %s to read a value for a char variable

Related

inputting a character string using scanf()

I started learning about inputting character strings in C. In the following source code I get a character array of length 5.
#include<stdio.h>
int main(void)
{
char s1[5];
printf("enter text:\n");
scanf("%s",s1);
printf("\n%s\n",s1);
return 0;
}
when the input is:
1234567891234567, and I've checked it's working fine up to 16 elements(which I don't understand because it is more than 5 elements).
12345678912345678, it's giving me an error segmentation fault: 11 (I gave 17 elements in this case)
123456789123456789, the error is Illegal instruction: 4 (I gave 18 elements in this case)
I don't understand why there are different errors. Is this the behavior of scanf() or character arrays in C?. The book that I am reading didn't have a clear explanation about these things. FYI I don't know anything about pointers. Any further explanation about this would be really helpful.
Is this the behavior of scanf() or character arrays in C?
TL;DR - No, you're facing the side-effects of undefined behavior.
To elaborate, in your case, against a code like
scanf("%s",s1);
where you have defined
char s1[5];
inputting anything more than 4 char will cause your program to venture into invalid memory area (past the allocated memory) which in turn invokes undefined behavior.
Once you hit UB, the behavior of the program cannot be predicted or justified in any way. It can do absolutely anything possible (or even impossible).
There is nothing inherent in the scanf() which stops you from reading overly long input and overrun the buffer, you should keep control on the input string scanning by using the field width, like
scanf("%4s",s1); //1 saved for terminating null
The scanf function when reading strings read up to the next white-space (e.g. newline, space, tab etc.), or the "end of file". It has no idea about the size of the buffer you provide it.
If the string you read is longer than the buffer provided, then it will write out of bounds, and you will have undefined behavior.
The simplest way to stop this is to provide a field length to the scanf format, as in
char s1[5];
scanf("%4s",s1);
Note that I use 4 as field length, as there needs to be space for the string terminator as well.
You can also use the "secure" scanf_s for which you need to provide the buffer size as an argument:
char s1[5];
scanf_s("%s", s1, sizeof(s1));

What happens when a scanf("%s",s); encounters two or more words and only one variable is provided?

When a scanf("%s",s); (one of many ways to get a string which is not perfect) encounters a space in the input, it will try to put it in another variable, right? But what hapens if there is only one variable provided as in this case?
Also what other ways are used to input a string? which is the esiest or best one to use and which one does not give problems like the gets(s); function?
Here is my s_insert function now:
// pointer to pointer needed when you allocate memory in the function
void s_insert(char **string_one){ //inserts string (loss of original data)
*string_one=(char*)malloc(200);
fgets (*string_one,200,stdin);
}
scanf for %s data specification reads characters before first space symbol (' ', '\n' or '\t'). If you want to read string with spaces (more than two words) use fgets function, that is more safe than 'gets' because you can set the maximum number of character that can be allocated in your memory and avoid segmentation fault.
No, it will only try to "put in a variable" when it encounters a % with a suitable conversion specifier in the first argument. That argument is what controls its behavior, not the input.
Having a single %s and multiple words in the input will simply leave the remaining words still in the input buffer, since scanf() will stop when it's done with the single %s, it has nothing more to do then.
It reads its conversion specification string and tries to read input to match that, not the other way around.

Why output length is coming 6?

I have written a simple program to calculate length of string in this way.
I know that there are other ways too. But I just want to know why this program is giving this output.
#include <stdio.h>
int main()
{
char str[1];
printf( "%d", printf("%s", gets(str)));
return 0;
}
OUTPUT :
(null)6
Unless you always pass empty strings from the standard input, you are invoking undefined behavior, so the output could be pretty much anything, and it could crash as well. str cannot be a well-formed C string of more than zero characters.
char str[1] allocates storage room for one single character, but that character needs to be the NUL character to satisfy C string constraints. You need to create a character array large enough to hold the string that you're writing with gets.
"(null)6" as the output could mean that gets returned NULL because it failed for some reason or that the stack was corrupted in such a way that the return value was overwritten with zeroes (per the undefined behavior explanation). 6 following "(null)" is expected, as the return value of printf is the number of characters that were printed, and "(null)" is six characters long.
There's several issues with your program.
First off, you're defining a char buffer way too short, a 1 char buffer for a string can only hold one string, the empty one. This is because you need a null at the end of the string to terminate it.
Next, you're using the gets function which is very unsafe, (as your compiler almost certainly warned you about), as it just blindly takes input and copies it into a buffer. As your buffer is 0+terminator characters long, you're going to be automatically overwriting the end of your string into other areas of memory which could and probably does contain important information, such as your rsp (your return pointer). This is the classic method of smashing the stack.
Third, you're passing the output of a printf function to another printf. printf isn't designed for formating strings and returning strings, there are other functions for that. Generally the one you will want to use is sprintf and pass it in a string.
Please read the documentation on this sort of thing, and if you're unsure about any specific thing read up on it before just trying to program it in. You seem confused on the basic usage of many important C functions.
It invokes undefined behavior. In this case you may get any thing. At least str should be of 2 bytes if you are not passing a empty string.
When you declare a variable some space is reserved to store the value.
The reserved space can be a space that was previously used by some other
code and has values. When the variable goes out of scope or is freed
the value is not erased (or it may be, anything goes.) only the programs access
to that variable is revoked.
When you read from an unitialised location you can get anything.
This is undefined behaviour and you are doing that,
Output on gcc (Ubuntu/Linaro 4.6.3-1ubuntu5) 4.6.3 is 0
For above program your input is "(null)", So you are getting "(null)6". Here "6" is the output from printf (number of characters successfully printed).

Why does C's printf format string have both %c and %s?

Why does C's printf format string have both %c and %s?
I know that %c represents a single character and %s represents a null-terminated string of characters, but wouldn't the string representation alone be enough?
Probably to distinguish between null terminated string and a character. If they just had %s, then every single character must also be null terminated.
char c = 'a';
In the above case, c must be null terminated. This is my assumption though :)
%s prints out chars until it reaches a 0 (or '\0', same thing).
If you just have a char x;, printing it with printf("%s", &x); - you'd have to provide the address, since %s expects a char* - would yield unexpected results, as &x + 1 might not be 0.
So you couldn't just print a single character unless it was null-terminated (very inefficent).
EDIT: As other have pointed out, the two expect different things in the var args parameters - one a pointer, the other a single char. But that difference is somewhat clear.
The issue that is mentioned by others that a single character would have to be null terminated isn't a real one. This could be dealt with by providing a precision to the format %.1s would do the trick.
What is more important in my view is that for %s in any of its forms you'd have to provide a pointer to one or several characters. That would mean that you wouldn't be able to print rvalues (computed expressions, function returns etc) or register variables.
Edit: I am really pissed off by the reaction to this answer, so I will probably delete this, this is really not worth it. It seems that people react on this without even having read the question or knowing how to appreciate the technicality of the question.
To make that clear: I don't say that you should prefer %.1s over %c. I only say that reasons why %c cannot be replaced by that are different than the other answer pretend to tell. These other answers are just technically wrong. Null termination is not an issue with %s.
The printf function is a variadic function, meaning that it has variable number of arguments. Arguments are pushed on the stack before the function (printf) is called. In order for the function printf to use the stack, it needs to know information about what is in the stack, the format string is used for that purpose.
e.g.
printf( "%c", ch ); tells the function the argument 'ch'
is to be interpreted as a character and sizeof(char)
whereas
printf( "%s", s ); tells the function the argument 's' is a pointer
to a null terminated string sizeof(char*)
it is not possible inside the printf function to otherwise determine stack contents e.g. distinguishing between 'ch' and 's' because in C there is no type checking during runtime.
%s says print all the characters until you find a null (treat the variable as a pointer).
%c says print just one character (treat the variable as a character code)
Using %s for a character doesn't work because the character is going to be treated like a pointer, then it's going to try to print all the characters following that place in memory until it finds a null
Stealing from the other answers to explain it in a different way.
If you wanted to print a character using %s, you could use the following to properly pass it an address of a char and to keep it from writing garbage on the screen until finding a null.
char c = 'c';
printf('%.1s', &c);
For %s, we need provide the address of string, not its value.
For %c, we provide the value of characters.
If we used the %s instead of %c, how would we provide a '\0' after the characters?
Id like to add another point of perspective to this fun question.
Really this comes down to data typing. I have seen answers on here that state that you could provide a pointer to the char, and provide a
"%.1s"
This could indeed be true. But the answer lies in the C designer's trying to provide flexibility to the programmer, and indeed a (albeit small) way of decreasing footprint of your application.
Sometimes a programmer might like to run a series of if-else statements or a switch-case, where the need is to simply output a character based upon the state. For this, hard coding the the characters could indeed take less actual space in memory as the single characters are 8 bits versus the pointer which is 32 or 64 bits (for 64 bit computers). A pointer will take up more space in memory.
If you would like to decrease the size through using actual chars versus pointers to chars, then there are two ways one could think to do this within printf types of operators. One would be to key off of the .1s, but how is the routine supposed to know for certain that you are truly providing a char type versus a pointer to a char or pointer to a string (array of chars)? This is why they went with the "%c", as it is different.
Fun Question :-)
C has the %c and %s format specifiers because they handle different types.
A char and a string are about as different as night and 1.
%c expects a char, which is an integer value and prints it according to encoding rules.
%s expects a pointer to a location of memory that contains char values, and prints the characters in that location according to encoding rules until it finds a 0 (null) character.
So you see, under the hood, the two cases while they look alike they have not much in common, as one works with values and the other with pointers. One is instructions for interpreting a specific integer value as an ascii char, and the other is iterating the contents of a memory location char by char and interpreting them until a zero value is encountered.
I have done a experiment with printf("%.1s", &c) and printf("%c", c).
I used the code below to test, and the bash's time utility the get the runing time.
#include<stdio.h>
int main(){
char c = 'a';
int i;
for(i = 0; i < 40000000; i++){
//printf("%.1s", &c); get a result of 4.3s
//printf("%c", c); get a result of 0.67s
}
return 0;
}
The result says that using %c is 10 times faster than %.1s. So, althought %s can do the job of %c, %c is still needed for performance.
Since no one has provided an answer with ANY reference whatsoever, here is a printf specification from pubs.opengroup.com which is similar to the format definition from IBM
%c
The int argument shall be converted to an unsigned char, and the resulting byte shall be written.
%s
The argument shall be a pointer to an array of char. Bytes from the array shall be written up to (but not including) any terminating null byte. If the precision is specified, no more than that many bytes shall be written. If the precision is not specified or is greater than the size of the array, the application shall ensure that the array contains a null byte.

Using scanf to read in certain amount of characters in C?

I am having trouble accepting input from a text file. My program is supposed to read in a string specified by the user and the length of that string is determined at runtime. It works fine when the user is running the program (manually inputting the values) but when I run my teacher's text file, it runs into an infinite loop.
For this example, it fails when I am taking in 4 characters and his input in his file is "ABCDy". "ABCD" is what I am supposed to be reading in and 'y' is supposed to be used later to know that I should restart the game. Instead when I used scanf to read in "ABCD", it also reads in the 'y'. Is there a way to get around this using scanf, assuming I won't know how long the string should be until runtime?
Normally, you'd use something like "%4c" or "%4s" to read a maximum of 4 characters (the difference is that "%4c" reads the next 4 characters, regardless, while "%4s" skips leading whitespace and stops at a whitespace if there is one).
To specify the length at run-time, however, you have to get a bit trickier since you can't use a string literal with "4" embedded in it. One alternative is to use sprintf to create the string you'll pass to scanf:
char buffer[128];
sprintf(buffer, "%%%dc", max_length);
scanf(buffer, your_string);
I should probably add: with printf you can specify the width or precision of a field dynamically by putting an asterisk (*) in the format string, and passing a variable in the appropriate position to specify the width/precision:
int width = 10;
int precision = 7;
double value = 12.345678910;
printf("%*.*f", width, precision, value);
Given that printf and scanf format strings are quite similar, one might think the same would work with scanf. Unfortunately, this is not the case--with scanf an asterisk in the conversion specification indicates a value that should be scanned, but not converted. That is to say, something that must be present in the input, but its value won't be placed in any variable.
Try
scanf("%4s", str)
You can also use fread, where you can set a read limit:
char string[5]={0};
if( fread(string,(sizeof string)-1,1,stdin) )
printf("\nfull readed: %s",string);
else
puts("error");
You might consider simply looping over calls to getc().

Resources