I am trying to understand string's behavior in C and it is bothering me since my following two code snippets result into different output:
(For the sake of this question, Let us assume user enters 12)
int main(void)
{
char L_Red[2];
char temp[] = "I";
printf("Enter pin connected to red: ");
scanf("%s", L_Red);
strcat(temp,L_Red);
printf("%s \n", temp);
return 0;
}
this yields: 12 as output (and not I12) Why ?
int main(void)
{
char L_Red[2];
printf("Enter pin connected to red: ");
scanf("%s", L_Red);
char temp[] = "I";
strcat(temp,L_Red);
printf("%s \n", temp);
return 0;
}
This yields: I12I (and not, I12) Why ?
I have read about string in C and as per my understanding, neither am I allocating temp any fixed size and changing it later to get these vague outputs nor am I using strings like the way they are not supposed to. Is there any other concept at play here ?
The array temp is an array of two characters (the 'I' and the string terminator '\0'). That's it. Attempting to append more characters to that array will write out of bounds and lead to undefined behavior.
You need to make sure that the destination array temp have enough space to fit its original content plus the string you want to append (plus the terminator).
Also, if you want to input more than one character for the "string" L_Red you need to increase its size as well.
I also recommend you use a limit in the format specifier so you can't write out of bounds:
char L_Red[3]; // Space for two characters, plus terminator
scanf("%2s", L_Red); // Read at most two characters of input
You are getting strange answers because your destination string (ie the first argument to strcat) is not long enough to handle both strings plus a null termination character. Also the length of L_Red is too short as it does not have enough space for the null termination character either.
Related
For an instance if I store ABCDE from scanf function, the later printf function gives me ABCDE as output. So what is the point of assigning the size of the string(Here 4).
#include <stdio.h>
int main() {
int c[4];
printf("Enter your name:");
scanf("%s",c);
printf("Your Name is:%s",c);
return 0;
}
I'll start with, don't use int array to store strings!
int c[4] allocates an array of 4 integers. An int is typically 4 bytes, so usually this would be 16 bytes (but might be 8 or 32 or something else on some platforms).
Then, you use this allocation first to read characters with scanf. If you enter ABCDE, it uses up 6 characters (there is an extra 0 byte at the end of the string marking the end, which needs space too), which happens to fit into the memory reserved for array of 4 integers. Now you could be really unlucky and have a platform where int has a so called "trap representation", which would cause your program to crash. But, if you are not writing the code for some very exotic device, there won't be. Now it just so happens, that this code is going to work, for the same reason memcpy is going to work: char type is special in C, and allows copying bytes to and from different types.
Same special treatment happens, when you print the int[4] array with printf using %s format. It works, because char is special.
This also demonstrates how very unsafe scanf and printf are. They happily accept c you give them, and assume it is a char array with valid size and data.
But, don't do this. If you want to store a string, use char array. Correct code for this would be:
#include <stdio.h>
int main() {
char c[16]; // fits 15 characters plus terminating 0
printf("Enter your name:");
int items = scanf("%15s",c); // note: added maximum characters
// scanf returns number of items read successfully, *always* check that!
if (items != 1) {
return 1; // exit with error, maybe add printing error message
}
printf("Your Name is: %s\n",c); // note added newline, just as an example
return 0;
}
The size of an array must be defined while declaring a C String variable because it is used to calculate how many characters are going to be stored inside the string variable and thus how much memory will be reserved for your string. If you exceed that amount the result is undefined behavior.
You have used int c , not char c . In C, a char is only 1 byte long, while a int is 4 bytes. That's why you didn't face any issues.
(Simplifying a fair amount)
When you initialize that array of length 4, C goes and finds a free spot in memory that has enough consecutive space to store 4 integers. But if you try to set c[4] to something, C will write that thing in the memory just after your array. Who knows what’s there? That might not be free, so you might be overwriting something important (generally bad). Also, if you do some stuff, and then come back, something else might’ve used that memory slot (properly) and overwritten your data, replacing it with bizarre, unrelated, and useless (to you) data.
In C language the last of the string is '\0'.
If you print with the below function, you can see the last character of the string.
scanf("%s", c); add the last character, '\0'.
So, if you use another function, getc, getch .., you should consider adding the laster character by yourself.
#include<stdio.h>
#include<string.h>
int main(){
char c[4+1]; // You should add +1 for the '\0' character.
char *p;
int len;
printf("Enter your name:");
scanf("%s", c);
len = strlen(c);
printf("Your Name is:%s (%d)\n", c, len);
p = c;
do {
printf("%x\n", *(p++));
} while((len--)+1);
return 0;
}
Enter your name:1234
Your Name is:1234 (4)
31
32
33
34
0 --> last character added by scanf("%s);
ffffffae --> garbage
I have a simple error that I know lies underneath my C code's memory usage, because the iterator in a simple for loop changes drastically after receiving user input on the command line:
int i = 1;
char input[] = "";
for (i = 1; i <= 5; i++) {
printf("i %d\n> ", i);
scanf("%s", input);
printf("input %s\ni %d\n", input, i);
}
The output should be simple enough:
i 1
> <receive input>
input <input>
i 1
to be repeated 5 times.
However, the iterator 'i' changes to anything but what is expected when any input is received.
An example output:
i 1
> 45
input 45
i 53
I have a hunch that this comes from memory access in the compiler. Anything helps!
scanf("%s", input);
From the docs of scanf:
%s Matches a sequence of bytes that are not white-space characters. The application shall ensure that the corresponding argument is a pointer to the initial byte of an array of char, signed char, or unsigned char large enough to accept the sequence and a terminating null character code, which shall be added automatically.
You are the application. char input[] = ""; is only 1 byte big.
Any character returned by the scanf will result in overflowing input due to the null terminator. And will write over the next variable in memory.
Try:
char input[100] = "";
scanf("%100s", input);
Look at how your local variables are declared:
int i = 1;
char input[] = "";
input is a zero-length string, and there's no room allocated for the input you're about to ask for. When you do:
scanf("%s", input);
the input gets written into the array pointed to by input, but since there was no space reserved for that, whatever happens to be after the array that input refers to gets written over. In this case, that's I.
To solve the problem, you need to make sure that there's enough room for the input at the location where you're putting it. Also, you should limit the allowable length of the input so that the user can't enter more data than the size of the space you've reserved.
I'm new to C and I'm trying to write a program that prints the ASCII value for every letter in a name that the user enters. I attempted to store the letters in an array and try to print each ASCII value and letter of the name separately but, for some reason, it only prints the value of the first letter.
For example, if I write "Anna" it just prints 65 and not the values for the other letters in the name. I think it has something to do with my sizeof(name)/sizeof(char) part of the for loop, because when I print it separately, it only prints out 1.
I can't figure out how to fix it:
#include <stdio.h>
int main(){
int e;
char name[] = "";
printf("Enter a name : \n");
scanf("%c",&name);
for(int i = 0; i < (sizeof(name)/sizeof(char)); i++){
e = name[i];
printf("The ASCII value of the letter %c is : %d \n",name[i],e);
}
int n = (sizeof(name)/sizeof(char));
printf("%d", n);
}
Here's a corrected, annotated version:
#include <stdio.h>
#include <string.h>
int main() {
int e;
char name[100] = ""; // Allow for up to 100 characters
printf("Enter a name : \n");
// scanf("%c", &name); // %c reads a single character
scanf("%99s", name); // Use %s to read a string! %99s to limit input size!
// for (int i = 0; i < (sizeof(name) / sizeof(char)); i++) { // sizeof(name) / sizeof(char) is a fixed value!
size_t len = strlen(name); // Use this library function to get string length
for (size_t i = 0; i < len; i++) { // Saves calculating each time!
e = name[i];
printf("The ASCII value of the letter %c is : %d \n", name[i], e);
}
printf("\n Name length = %zu\n", strlen(name)); // Given length!
int n = (sizeof(name) / sizeof(char)); // As noted above, this will be ...
printf("%d", n); // ... a fixed value (100, as it stands).
return 0; // ALWAYS return an integer from main!
}
But also read the comments given in your question!
This is a rather long answer, feel free to skip to the end for the code example.
First of all, by initialising a char array with unspecified length, you are making that array have length 1 (it only contains the empty string). The key issue here is that arrays in C are fixed size, so name will not grow larger.
Second, the format specifier %c causes scanf to only ever read one byte. This means that even if you had made a larger array, you would only be reading one byte to it anyway.
The parameter you're giving to scanf is erroneous, but accidentally works - you're passing a pointer to an array when it expects a pointer to char. It works because the pointer to the array points at the first element of the array. Luckily this is an easy fix, an array of a type can be passed to a function expecting a pointer to that type - it is said to "decay" to a pointer. So you could just pass name instead.
As a result of these two actions, you now have a situation where name is of length 1, and you have read exactly one byte into it. The next issue is sizeof(name)/sizeof(char) - this will always equal 1 in your program. sizeof char is defined to always equal 1, so using it as a divisor causes no effect, and we already know sizeof name is equal to 1. This means your for loop will only ever read one byte from the array. For the exact same reason n is equal to 1. This is not erroneous per se, it's just probably not what you expected.
The solution to this can be done in a couple of ways, but I'll show one. First of all, you don't want to initialize name as you do, because it always creates an array of size 1. Instead you want to manually specify a larger size for the array, for instance 100 bytes (of which the last one will be dedicated to the terminating null byte).
char name[100];
/* You might want to zero out the array too by eg. using memset. It's not
necessary in this case, but arrays are allowed to contain anything unless
and until you replace their contents.
Parameters are target, byte to fill it with, and amount of bytes to fill */
memset(name, 0, sizeof(name));
Second, you don't necessarily want to use scanf at all if you're reading just a byte string from standard input instead of a more complex formatted string. You could eg. use fgets to read an entire line from standard input, though that also includes the newline character, which we'll have to strip.
/* The parameters are target to write to, bytes to write, and file to read from.
fgets writes a null terminator automatically after the string, so we will
read at most sizeof(name) - 1 bytes.
*/
fgets(name, sizeof(name), stdin);
Now you've read the name to memory. But the size of name the array hasn't changed, so if you used the rest of the code as is you would get a lot of messages saying The ASCII value of the letter is : 0. To get the meaningful length of the string, we'll use strlen.
NOTE: strlen is generally unsafe to use on arbitrary strings that might not be properly null-terminated as it will keep reading until it finds a zero byte, but we only get a portable bounds-checked version strnlen_s in C11. In this case we also know that the string is null-terminated, because fgets deals with that.
/* size_t is a large, unsigned integer type big enough to contain the
theoretical maximum size of an object, so size functions often return
size_t.
strlen counts the amount of bytes before the first null (0) byte */
size_t n = strlen(name);
Now that we have the length of the string, we can check if the last byte is the newline character, and remove it if so.
/* Assuming every line ends with a newline, we can simply zero out the last
byte if it's '\n' */
if (name[n - 1] == '\n') {
name[n - 1] = '\0';
/* The string is now 1 byte shorter, because we removed the newline.
We don't need to calculate strlen again, we can just do it manually. */
--n;
}
The loop looks quite similar, as it was mostly fine to begin with. Mostly, we want to avoid issues that can arise from comparing a signed int and an unsigned size_t, so we'll also make i be type size_t.
for (size_t i = 0; i < n; i++) {
int e = name[i];
printf("The ASCII value of the letter %c is : %d \n", name[i], e);
}
Putting it all together, we get
#include <stdio.h>
#include <string.h>
int main() {
char name[100];
memset(name, 0, sizeof(name));
printf("Enter a name : \n");
fgets(name, sizeof(name), stdin);
size_t n = strlen(name);
if (n > 0 && name[n - 1] == '\n') {
name[n - 1] = '\0';
--n;
}
for (size_t i = 0; i < n; i++){
int e = name[i];
printf("The ASCII value of the letter %c is : %d \n", name[i], e);
}
/* To correctly print a size_t, use %zu */
printf("%zu\n", n);
/* In C99 main implicitly returns 0 if you don't add a return value
yourself, but it's a good habit to remember to return from functions. */
return 0;
}
Which should work pretty much as expected.
Additional notes:
This code should be valid C99, but I believe it's not valid C89. If you need to write to the older standard, there are several things you need to do differently. Fortunately, your compiler should warn you about those issues if you tell it which standard you want to use. C99 is probably the default these days, but older code still exists.
It's a bit inflexible to be reading strings into fixed-size buffers like this, so in a real situation you might want to have a way of dynamically increasing the size of the buffer as necessary. This will probably require you to use C's manual memory management functionality like malloc and realloc, which aren't particularly difficult but take greater care to avoid issues like memory leaks.
It's not guaranteed the strings you're reading are in any specific encoding, and C strings aren't really ideal for handling text that isn't encoded in a single-byte encoding. There is support for "wide character strings" but probably more often you'll be handling char strings containing UTF-8 where a single codepoint might be multiple bytes, and might not even represent an individual letter as such. In a more general-purpose program, you should keep this in mind.
If we need write a code to get ASCII values of all elements in a string, then we need to use "%d" instead of "%c". By doing this %d takes the corresponding ascii value of the following character.
If we need to only print the ascii value of each character in the string. Then this code will work:
#include <stdio.h>
char str[100];
int x;
int main(){
scanf("%s",str);
for(x=0;str[x]!='\0';x++){
printf("%d\n",str[x]);
}
}
To store all corresponding ASCII value of character in a new variable, we need to declare an integer variable and assign it to character. By this way the integer variable stores ascii value of character. The code is:
#include <stdio.h>
char str[100];
int x,ascii;
int main(){
scanf("%s",str);
for(x=0;str[x]!='\0';x++){
ascii=str[x];
printf("%d\n",ascii);
}
}
I hope this answer helped you.....😊
I compiled this code using gcc (tdm-1) 5.1.0 and please tell me why the output doesn't contain "hello"
#include<stdio.h>
void main()
{
int i;
char st[20];
printf("Enter a string ");
scanf("%s",st);
for(i=0;i<20;i++)
{
printf("%c",st[i]);
}
}
Input:hello
Output: # #
You print all 20 elements of the array, but if the user entered a string smaller than that not all elements would be initialized. They would be indeterminate and seemingly random.
Remember that char strings in C are really called null-terminated byte strings. That null-terminated bit is important, and mean you can easily find the end of the string by checking the current character agains '\0' (which is the terminator character).
Or you could just use the strlen function to get the length of the string instead:
for(i=0;i<strlen(st);i++) { ... }
Or use the "%s" format to print the string:
printf("%s", st);
Also note that without any protection the scanf function will allow you give longer input than is space for in the array, so you need to protect agains that, for example by limiting the amount of characters scanf will read:
scanf("%19s",st); // Write at most 19 character (*plus* terminator) to the string
Now for why your input doesn't seem to be printed, it's because the indeterminate contents of the uninitialized elements. While you're not going out of bounds of your array, you still go out of bounds of the actual string. Going out of bounds leads to undefined behavior.
What's probably is happening is that some of the "random" indeterminate contents happens to be a carriage return '\r', which moves the cursor to the start of the line and the output already written will be overwritten by the uninitialized elements in your array.
Here's a short example as Qubit already explained:
#include <stdio.h>
void main () {
char str1[20];
printf("Enter name: ");
scanf("%s", str1);
printf("Entered Name: %s", str1);
}
Here
char st[20];
st is a local variable & default array st contents are garbage not zero. So if you scan less than 20 characters into st, in that case remaining location of array st contains garbage, hence it's printing some junk data like # # in case of
char st[20];
printf("Enter a string ");
scanf("%s",st);
for(i=0;i<20;i++) {
printf("%c",st[i]);
}
& it's a bad practice as if user entered few char lets say 5 char, then your loop rotates 20 times, internally it will do more operations or consume more CPU cycle.
So if you want to print a char array char by char, then you should rotate a loop until \0 char encounters, for e.g
for(i=0;st[i];i++) { /* this fails when \0 encounters */
printf("%c",st[i]);
}
Or
as others suggested you can print char array st using single printf by using %s format specifier like
printf("%s\n",st); /*here printf starts printing from base address of st
and prints until \0 */
Also it's better to initialize char array st while declaring itself. for e.g
char st[20] ="";
#include <stdio.h>
int main(int argc, char *argv[])
{
int i, n, m;
scanf("%d %d", &n, &m);
char s[m][n];
for (i = 0; i < m; i++) {
printf("the string --\n");
scanf("%s", s[i]);
}
for (i = 0; i < m; i++) {
printf("the strings are %s \n",s[i]);
printf("\n");
}
return 0;
}
The output is:
2 2
the string --
10
the string --
11
the strings are 1011
the strings are 11
Why is the first string 1011 instead of 10?
In C, strings are represented as a sequence of char values, terminated by a null character (0 or '\0'). This means that to store a two-character string, you need space for three characters: the two characters of string content, plus the null terminator character.
Here, you've only allocated enough space for two characters in each string, but you need space for three.
So, it reads the first string into the array s[0], but the null terminator doesn't fit, and so it overflows into the second array s[1]. Now your array of arrays s looks like this: {{'1', '0'}, {'\0', ... }}.
Then, when it reads the second string into the array s[1], it overwrites the overflowed null terminator from before. And the null terminator for the second string doesn't fit into its own array, so it overflows again into the rest of the stack. The program might crash here, or corrupt other data, because you're overflowing past the end of the array.
So now your array of arrays s ends up looking like this: {{'1', '0'}, {'1', '1'}}, followed by a '\0' somewhere after the end of the array.
When printf goes to read your first string, it prints characters until it finds a null terminator. But it doesn't find one in the first string, so it keeps going, and hits the second string. It doesn't find one there either, and continues past the end of the array. In your case, luckily a null terminator was right there, but for all we know there could be something else.
To fix this, you need to allocate an extra character per string on line 9, for the null terminator:
char s[m][n+1];
There's another problem here, however. What if your input gives you the wrong length? For example, what if your input says 2 3, i.e. that the following strings will have a length of 3, but gives you the string foobar, which is 6 characters? Your code right now would overflow the buffer when it read that string, because it doesn't ensure it's the right length.
One way to avoid this would be to use gets_s instead of sscanf() for reading the strings on line 13:
gets_s(s[i], n+1);
This will read at most n characters, so avoid crashing your program or creating a security issue. However, gets_s is a C11 function, so you may not be able to use it.
You must set column size to 3 if you insert 2 characters per string, 4 for 3 characters per string and so on.
This because string in C have a termination character ('\0') in the last position.
#include <stdio.h>
int main(int argc, char *argv[])
{
int i, n, m;
scanf("%d %d", &n, &m);
char s[m][n+1];
for (i = 0; i < m; i++) {
printf("the string --\n");
scanf("%s", s[i]);
}
for (i = 0; i < m; i++) {
printf("the strings are %s \n",s[i]);
printf("\n");
}
return 0;
}
%s expects null terminated strings as an argument. When first string is read by scanf, there is not enough space for the null terminator within the allocated memory of first string. It will goes to the space next to the allocated space. Writing to unallocated space invokes undefined behavior.
While printing the strings with %s specifier, printf write the string character by character till it finds a null terminator '\0'. Here it may be the case that both the strings 10 and 11 are stored one after another in memory, so printf writes the first string till it read the null character of second string.
Input n as 3 and you will get the desire results.
In my opninion using scanf to read strings is just pure evil.
That said the array s[m][n] is just s[m*n] of course.
That said that evil thing scanf is going to load on *s[0] 10\n\0 and on *s[1*n] or *s[2] 11 and *s will be 1011\n\0
And this is a monument to bad C coding. I guess it's just an example but if I was asked this question i would say: "Come on, get me real things"