I am having problems with the printf function in the CS50 IDE. When I am using printf to print out a string (salt in this code), extra characters are being output that were not present in the original argument (argv).
Posted below is my code. Any help would be appreciated. Thank you.
#include <cs50.h>
#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <stdlib.h>
int main(int argc, string argv[])
{
// ensuring that only 1 command-line argument is inputted
if (argc != 2)
{
return 1;
}
char salt[2];
for (int i = 0; i < 2; i++)
{
char c = argv[1][i];
salt[i] = c;
}
printf("the first 2 characters of the argument is %s\n", salt);
}
You are missing a string terminator in salt.
Somehow the computer needs to know where your string ends in memory. It does so by reading until it encounters a NUL byte, which is a byte with value zero.
Your array salt has exactly 2 bytes of space, and after them, random garbage exists which just happens to be next in memory after your array. Since you don't have a string terminator, the computer will read this garbage as well until it encounters a NUL byte.
All you need to do is include such a byte in your array, like so:
char salt[3] = {0};
This will make salt one byte longer, and the {0} is a shorthand for {0, 0, 0} which will initialize the contents of the array with all zerores. (Alternatively, you could use char salt[3]; and later manually set the last byte to zero using salt[2] = 0;.)
In your case, salt is at least one element shy of being a string, unless the argv[1] is only one element, it does not contain a null-terminator.
You need to allocate space to hold the null-terminator and actually put one there to be able to use salt as string, as expected for the argument to %s conversion specifier in case of printf().
Otherwise, the string related functions and operations, which essentially rely on the fact that there will be a null terminator to mark the end of the char array (i.e., mark the end of valid memory that can be accessed), will try to access past the valid memory which causes undefined behavior. Once you hit UB, nothing is guaranteed.
So, considering the fact that you want to use
"....the first 2 characters of the argument....."
you need to make salt a 3-element char array, and make sure that salt[2] contains a null-terminator, like '\0'.
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'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'm relatively new to C. I wanted to lern the language a bit by solving coderbyte challenges.
But I'm stucked at the first. It is supposed to be a simple String reverse algorithm.
When I input things like "asdf" or "1234567" the output is correct ("fdsa", "7654321"). But when I type "12345678" or "thisiscool" I get "87654321▒#"/"loocsisiht#" as a result. I don't know where the # are comming from.
This is my code:
#include <stdio.h>
#include <string.h>
void FirstReverse(char str[]) {
int len = strlen(str);
char nstr[len];
int i;
for(i = 0; i < len; i++) {
nstr[i] = *(str+len-1-i);
}
printf("%s\n", nstr);
}
int main(void) {
char str[100];
FirstReverse(gets(str));
return 0;
}
Can someone please tell me where I can find the error?
Thanks in advance :)
In C, strings are zero-terminated. A string "cat", for example, has 4 characters, and is represented as ('c','a','t',(char)0). You forgot about the final 0.
Note that strlen returns the string length without the final 0, so a string foo contains strlen(foo)+1 characters. Remember this when you allocate strings.
As the other answers have mentioned, you're missing a terminator.
It should also be noted that it's bad practice to allocate strings the way you did. An array should always have a fixed size if you create it that way.
You should instead do:
char * nstr = malloc(sizeof(char) * (len+1));
Thereby allocating the size of each character (1 byte) times the lenght.
Note the +1 because you need room for the string terminator.
When you call printf(, string); , it's gonna start from the first letter and print everything up to the terminator. Since you have no terminator here, it prints random characters, such as #.
What you're gonna wanna do to fix that, is adding:
nstr[i] = '\0';
after your loop.
Also remember to free the allocated memory.
You forgot to allocate a char for the terminating '\0' in nstr[].
So, better use: char nstr[len + 1]; and set nstr[len] = 0;
Furthermore: gets() is evil: from the glibc manual page:
Never use gets(). Because it is impossible to tell without knowing the data in advance how many characters gets() will read, and because gets() will continue to store characters past the end of the buffer, it is extremely dangerous to use. It has been used to break computer security. Use fgets() instead.
I would like to parse a specific line. So, I wrote the following piece of code in order to test the logic but I probably understand something wrongly :
typedef struct vers
{
char tu8UVersion[5];
char tu8UCommit[32];
}tst_prg_versions;
int main(int argc, char **argv)
{
tst_prg_versions lstVer;
char buf1[32];
char buf2[32];
char str[] = "BOARD-VERS-v1.0.0-git+9abc12345a";
sscanf(str, "BOARD-VERS-v%5s-git+%s", lstVer.tu8UVersion, lstVer.tu8UCommit);
printf("vers='%s'\n", lstVer.tu8UVersion);
printf("commit='%s'\n", lstVer.tu8UCommit);
sscanf(str, "BOARD-VERS-v%5s-git+%s", buf1, buf2);
printf("vers='%s'\n", buf1);
printf("commit='%s'\n", buf2);
return 0;
}
Once executed it returns :
vers='1.0.09abc12345a'
commit='9abc12345a'
vers='1.0.0'
commit='9abc12345a
Why the first vers is equals to 1.0.09abc12345a and not 1.0.0 ?
The first actually reads 1.0.0! Problem is, however, that tu8UVersion is not null-terminated, thus printf (not sscanf) prints beyound the field (doing so is undefined behaviour, however, as noted by sjsam) - which is immediately followed by tu8UCommit (does not necessarily have to be so, there could still be some fill bytes in between for alignment reasons!).
You need to either print 5 characters at most (%.5s in printf format string) or leave place for terminating the tu8UVersion with 0, as proposed in a comment already.
Something similar could have happened with your buffers, too. You are lucky that they appearently have been initialized to 0 already (probably because of compiled as debug version), which again does not necessarily have to happen. So with bad luck, you could have printed the whole rest of buf1 (having been left at garbage) and even beyond.
Why the first vers is equals to 1.0.09abc12345a and not 1.0.0 ?
Remember that you have
typedef struct vers
{
char tu8UVersion[5];
char tu8UCommit[32];
}tst_prg_versions;
I guess, there is a good chance the memory for tu8UVersion and tu8UCommit is contiguous. Since you have not null-terminated tu8UVersion when you do :
printf("vers='%s'\n", lstVer.tu8UVersion);
it goes on to print tu8UCommit and it stops because tu8UCommit is null terminated.
While sscanf seem the most sensible solution here you could also introduce some formatting :
char tu8UVersion[32];
/* version number can't get too big.
* So the first step is do allocated a
* reasonably - but not too - big size for it.
* So that you can be sure there are few empty bytes at the end.
*/
and then use a function to sanitize a string :
char* sanitized(char* ptr)
{
if(ptr[strlen(ptr)]!='\0') // include string.h for strlen
ptr[strlen(ptr)]='\0';
return ptr;
}
and print it like :
printf("vers='%s'\n", sanitized(lstVer.tu8UVersion));
Your problem has already been identified in the comments: You don't leave space for the terminating null character and the two strings are run together.
If you want to scan a version whose size you don't know beforehand, you can limit the characters to scan to decimal digits and points with %[.-9] or to everything except a hyphen with %[^-]. (The %[...] format is like %s, except that you must provide a list of valid characters in the brackets. A caret as first letter means that the string is made up of characters that are not listed. In other words, %s is short for %[^ \t\n]
When you scan a string, you should test the return value of sscanf to be sure that all items have been scanned correctly and contain valid values.
Here's a variant that scans version numbers of up to 11 letters:
#include <stdlib.h>
#include <stdio.h>
typedef struct vers
{
char tu8UVersion[12];
char tu8UCommit[32];
} tst_prg_versions;
int main(int argc, char **argv)
{
tst_prg_versions lstVer;
char str[] = "BOARD-VERS-v1.0.0-git+9abc12345a";
int n;
n = sscanf(str, "BOARD-VERS-v%11[^-]-git+%s",
lstVer.tu8UVersion, lstVer.tu8UCommit);
if (n == 2) {
printf("vers='%s'\n", lstVer.tu8UVersion);
printf("commit='%s'\n", lstVer.tu8UCommit);
} else {
puts("Parse error.");
}
return 0;
}
Suppose i have array of characters. say char x[100]
Now, i take input from the user and store it in the char array. The user input is less than 100 characters. Now, if i want to do some operation on the valid values, how do i find how many valid values are there in the char array. Is there a C function or some way to find the actual length of valid values which will be less than 100 in this case.
Yes, C has function strlen() (from string.h), which gives you number of characters in char array. How does it know this? By definition, every C "string" must end with the null character. If it does not, you have no way of knowing how long the string is or with other words, values of which memory locations of the array are actually "useful" and which are just some dump. Knowing this, sizeof(your_string) returns the size of the array (in bytes) and NOT length of the string.
Luckily, most C library string functions that create "strings" or read input and store it into a char array will automatically attach null character at the end to terminate the "string". Some do not (for example strncpy() ). Be sure to read their descriptions carefully.
Also, take notice that this means that the buffer supplied must be at least one character longer than the specified input length. So, in your case, you must actually supply char array of length 101 to read in 100 characters (the difference of one byte is for the null character).
Example usage:
#include <stdio.h>
#include <string.h>
int main(void)
{
char *string = "Hello World";
printf("%lu\n", (unsigned long)strlen(string));
return 0;
}
strlen() is defined as:
size_t strlen(const char * str)
{
const char *s;
for (s = str; *s; ++s);
return(s - str);
}
As you see, the end of a string is found by searching for the first null character in the array.
That depends on entirely where you got the input. Most likely strlen will do the trick.
Every time you enter a string in array in ends with a null character. You just have to find where is the null character in array.
You can do this manually otherwise, strlen() will solve your problem.
char ch;
int len;
while( (ch=getche() ) != '13' )
{
len++;
}
or use strlen after converting from char to string by %s