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;
}
Related
int main(int argc, char* argv[]) {
char* string; //local variable for a character
string = (char*)malloc(sizeof(char));
char* yes = "yes";
printf("Do you want to play (y/n)?");
scanf_s("%s", string);
if (strcmp(yes, string) == 0) {
printf("Hello welcome...");
}
Above is my code. Essentially, I want to make a loop asking for the user to input yes, y to continue; n, no to discontinue. But I simplify it for the sake of simple codes. The program just output the question, I press enter yes and enter then it stops.
I cant figure out a way to do it using array syntax( char string[] way, although array and malloc are basically the same) so I use pointer and malloc instead.
I'm going mad because this is bugging me so much. The practice assignment only asks to input character 'y' 'n' using %c but i want to do it the %s.
Really appreciate any help, im really stuck now. Thank you so much
Your code has two significant problems. First, if you want to input a string of characters, your malloc call needs to allocate space for more than one character; you should allocate the maximum number of characters you think the user's input will contain plus one - strings in C have a zero (nul) character at the end, to mark the end of the string).
Second, when you use the scanf_s function to read in a string (using the %s format specifier, as you have done), then you need to add additional parameters (the size of the target string buffer) after each string argument.
Here's a modified version of your code with these (and a few other) corrections:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(int argc, char* argv[])
{
char* string = malloc(5); //local variable for a character string (up to 5 chars)
char* yes = "yes";
printf("Do you want to play (y/n)?");
scanf_s("%s", string, 5);
if (strcmp(yes, string) == 0) {
printf("Hello welcome...");
}
free(string); // Don't forget to release the memory!
return 0; // Conventionally, return 0 for success or non-zero on error
}
Note 1: Each and every call to malloc (or calloc) should be paired with a call to free to release the allocated memory, or you will end up with memory leaks.
Note 2: Please read this post: Do I cast the result of malloc?
Note 3: Although most (all?) C compilers will not insist on it, it is good practice to explicitly add a return 0; (for success) statement at the end of the main function.
Please feel free to ask for any further clarification and/or explanation.
I have tried this code to separate my Str[] string into 2 string, but my problem is "I want to separate John(name) as string and 100(marks) as integer",How can I do it, any suggestion?
#include <stdio.h>
#include <string.h>
void main()
{
char Str[] = "John,100";
int i, j, xchange;
char name[50];
char marks[10];
j = 0; xchange = 0;
for(i=0; Str[i]!='\0'; i++){
if(Str[i]!=',' && xchange!=-1){
name[i] = Str[i];
}else{
xchange = -1;
}
if(xchange==-1){
marks[j++] = Str[i+1];
}
}
printf("Student name is %s\n", name);
printf("Student marks is %s", marks);
}
How to separate "John,100" into 2 strings?
There are three common approaches:
Use strtok() to split the string into individual tokens. This will modify the original string, but is quite simple to implement:
int main(void)
{
char line[] = "John,100;passed";
char *name, *score, *status;
/* Detach the initial part of the line,
up to the first comma, and set name
to point to that part. */
name = strtok(line, ",");
/* Detach the next part of the line,
up to the next comma or semicolon,
setting score to point to that part. */
score = strtok(NULL, ",;");
/* Detach the final part of the line,
setting status to point to it. */
status = strtok(NULL, "");
Note that if you change char line[] = "John,100"; then status will be NULL, but the code is otherwise safe to run.
So, in practice, if you required all three fields to exist in line, it would be sufficient to ensure the last one was not NULL:
if (!status) {
fprintf(stderr, "line[] did not have three fields!\n");
return EXIT_FAILURE;
}
Use sscanf() to convert the string. For example,
char line[] = "John,100";
char name[20];
int score;
if (sscanf(line, "%19[^,],%d", name, &score) != 2) {
fprintf(stderr, "Cannot parse line[] correctly.\n");
return EXIT_FAILURE;
}
Here, the 19 refers to the number of chars in name (one is always reserved for the end-of-string nul char, '\0'), and [^,] is a string conversion, consuming everything except a comma. %d converts an int. The return value is the number of successful conversions.
This approach does not modify the original string, and it allows you to try a number of different parsing patterns; as long as you try them the most complex one first, you can allow multiple input formats with very little added code. I do this regularly when taking 2D or 3D vectors as inputs.
The downside is that sscanf() (all functions in the scanf family) ignores overflow. For example, on 32-bit architectures, the largest int is 2147483647, but scanf functions will happily convert e.g. 9999999999 to 1410065407 (or some other value!) without returning an error. You can only assume the numerical inputs are sane and within the limits; you cannot verify.
Use helper functions to tokenise and/or parse the string.
Typically, the helper functions are something like
char *parse_string(char *source, char **to);
char *parse_long(char *source, long *to);
where source is a pointer to the next character in the string to be parsed, and to is a pointer to where the parsed value will be stored; or
char *next_string(char **source);
long next_long(char **source);
where source is a pointer to a pointer to the next character in the string to be parsed, and the return value is the value of the extracted token.
These tend to be longer than above, and if written by me, then quite paranoid about the inputs they accept. (I want my programs to complain if their input cannot be reliably parsed, rather than silently produce garbage.)
If the data is some variant of CSV (comma-separated values) read from a file, then the proper approach is a different one: instead of reading the file line by line, you read the file token by token.
The only "trick" is to remember the separator character that ended the token (you can use ungetc() for this), and use a different function to (read and ignore the rest of the tokens in the current record, and) consume the newline separator.
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'.
My problem is when I try to save the string (series[0]) Inside (c[0])
and I display it, it always ignore the last digit.
For Example the value of (series[0]) = "1-620"
So I save this value inside (c[0])
and ask the program to display (c[0]), it displays "1-62" and ignores the last digit which is "0". How can I solve this?
This is my code:
#include <stdio.h>
int main(void)
{
int price[20],i=0,comic,j=0;
char name,id,book[20],els[20],*series[20],*c[20];
FILE *rent= fopen("read.txt","r");
while(!feof(rent))
{
fscanf(rent,"%s%s%s%d",&book[i],&els[i],&series[i],&price[i]);
printf("1.%s %s %s %d",&book[i],&els[i],&series[i],price[i]);
i++;
}
c[0]=series[0];
printf("\n%s",&c[0]);
return 0;
}
The use of fscanf and printf is wrong :
fscanf(rent,"%s%s%s%d",&book[i],&els[i],&series[i],&price[i]);
Should be:
fscanf(rent,"%c%c%s%d",&book[i],&els[i],series[i],&price[i]);
You have used the reference operator on a char pointer when scanf expecting a char pointer, also you read a string to book and else instead of one character.
printf("1.%s %s %s %d",&book[i],&els[i],&series[i],price[i]);
Should be:
printf("1.%c %c %s %d",book[i],els[i],series[i],price[i]);
And:
printf("\n%s",&c[0]);
Should be:
printf("\n%s",c[0]);
c is an array of char * so c[i] can point to a string and that is what you want to send to printf function.
*Keep in mind that you have to allocate (using malloc) a place in memory for all the strings you read before sending them to scanf:
e.g:
c[0] = (char*)malloc(sizeof(char)*lengthOfString+1);
and only after this you can read characters in to it.
or you can use a fixed size double character array:
c[10][20];
Now c is an array of 20 strings that can be up to 9 characters long.
Amongst other problems, at the end you have:
printf("\n%s",&c[0]);
There are multiple problems there. The serious one is that c[0] is a char *, so you're passing the address of a char * — a char ** — to printf() but the %s format expects a char *. The minor problem is that you should terminate lines of output with newline.
In general, you have a mess with your memory allocation. You haven't allocated space for char *series[20] pointers to point at, so you get undefined behaviour when you use it.
You need to make sure you've allocated enough space to store the data, and it is fairly clear that you have not done that. One minor difficulty is working out what the data looks like, but it seems to be a series of lines each with 3 words and 1 number. This code does that job a bit more reliably:
#include <stdio.h>
int main(void)
{
int price[20];
int i;
char book[20][32];
char els[20][32];
char series[20][20];
const char filename[] = "read.txt";
FILE *rent = fopen(filename, "r");
if (rent == 0)
{
fprintf(stderr, "Failed to open file '%s' for reading\n", filename);
return 1;
}
for (i = 0; i < 20; i++)
{
if (fscanf(rent, "%31s%31s%19s%d", book[i], els[i], series[i], &price[i]) != 4)
break;
printf("%d. %s %s %s %d\n", i, book[i], els[i], series[i], price[i]);
}
printf("%d titles read\n", i);
fclose(rent);
return 0;
}
There are endless ways this could be tweaked, but as written, it ensures no overflow of the buffers (by the counting loop and input conversion specifications including the length), detects when there is an I/O problem or EOF, and prints data with newlines at the end of the line. It checks and reports if it fails to open the file (including the name of the file — very important when the name isn't hard-coded and a good idea even when it is), and closes the file before exiting.
Since you didn't provide any data, I created some random data:
Tixrpsywuqpgdyc Yeiasuldknhxkghfpgvl 1-967 8944
Guxmuvtadlggwjvpwqpu Sosnaqwvrbvud 1-595 3536
Supdaltswctxrbaodmerben Oedxjwnwxlcvpwgwfiopmpavseirb 1-220 9698
Hujpaffaocnr Teagmuethvinxxvs 1-917 9742
Daojgyzfjwzvqjrpgp Vigudvipdlbjkqjm 1-424 4206
Sebuhzgsqpyidpquzjxswbccqbruqf Vuhssjvcjjylcevcisdzedkzlp 1-581 3451
Doeraxdmyqcbbzyp Litbetmttcgfldbhqqfdxqi 1-221 2485
Raqqctfdlhrmhtzusntvgbvotpk Iowdcqlwgljwlfvwhfmw 1-367 3505
Kooqkvabwemxoocjfaa Hicgkztiqvqdjjx 1-466 435
Lowywyzzkkrazfyjuggidsqfvzzqb Qiginniroivqymgseushahzlrywe 1-704 5514
The output from the code above on that data is:
0. Tixrpsywuqpgdyc Yeiasuldknhxkghfpgvl 1-967 8944
1. Guxmuvtadlggwjvpwqpu Sosnaqwvrbvud 1-595 3536
2. Supdaltswctxrbaodmerben Oedxjwnwxlcvpwgwfiopmpavseirb 1-220 9698
3. Hujpaffaocnr Teagmuethvinxxvs 1-917 9742
4. Daojgyzfjwzvqjrpgp Vigudvipdlbjkqjm 1-424 4206
5. Sebuhzgsqpyidpquzjxswbccqbruqf Vuhssjvcjjylcevcisdzedkzlp 1-581 3451
6. Doeraxdmyqcbbzyp Litbetmttcgfldbhqqfdxqi 1-221 2485
7. Raqqctfdlhrmhtzusntvgbvotpk Iowdcqlwgljwlfvwhfmw 1-367 3505
8. Kooqkvabwemxoocjfaa Hicgkztiqvqdjjx 1-466 435
9. Lowywyzzkkrazfyjuggidsqfvzzqb Qiginniroivqymgseushahzlrywe 1-704 5514
10 titles read
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I've been trying to learn C programming by reading a textbook, but am confused about how strings and substrings work.
I have an idea of what strings and substrings are from java, but can't figure out the syntax in C.
Here's a question from the book that I thought might be easy, but I can't get it.
Write and test a function hydroxide that returns a 1 for true if its string argument ends in the substring OH.
It recommends testing the function with KOH and NaCl.
Also, how would I remove and add letters at the end of the string?
Like, if for some reason I wanted to change NaCl to NaOH?
Any help and explanations would be really appreciated.
ETA:
I guess what I'm most confused on is how to make the program look at the last two letters in the string and compared them to OH.
I'm also not sure how to pass strings to functions.
String is a sequence of characters that ends with special null-terminated character '\0'. If there is no \0, functions that work with string won't stop until the \0 symbol is found. This character may happen in any place after the end of pseudo string (I mean a string without \0) and only then stop.
The following example shows the necessity of this null-terminated character:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char string[] = "Hello!";
printf("original string:\n%s\n\n", string);
memset(string, '-', 5);
printf("memset doesn't affect the last two symbols: '!' and '\\0':\n%s", string);
memset(string, '-', 6);
printf("\n\nmemset doesn't affect the last symbol: '\\0':\n%s\n\n", string);
memset(string, '-', 7);
printf("memset affects all symbols including null-terminated one:\n%s", string);
return 0;
}
/* OUTPUT:
original string:
Hello!
memset doesn't affect the last two characters: '!' and '\0':
-----!
memset doesn't affect the last character: '\0':
------
memset affects all characters including null-terminated one:
-------#↓#
*/
Substring is a char sequence that is in a string. It may be less or equal to the string.
Suppose, "NaOH" is a string. Then substring may be: "N", "a", "O", "H", "Na", "aO", "OH", "NaO", "aOH", "NaOH". To find whether substring is in the string or not you can use strstr function. It's prototype is char * strstr ( char * str1, const char * str2 );.
This code shows this function's results:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char *ptrCh = NULL;
ptrCh = strstr("hello", "h");
printf("ptrCh: %p\n", ptrCh);
printf("%s\n\n", ptrCh);
ptrCh = strstr("hello", "z");
printf("ptrCh: %p\n", ptrCh);
printf("%s\n\n", ptrCh);
return 0;
}
/* OUTPUT:
ptrCh: 00403024
hello
ptrCh: 00000000
(null)
*/
As for the first printf, it prints characters beginning from the position of 'h' and when it reaches null-terminated character, which is next after 'o', it stops, exactly as in previous program.
To make your program more interactive, you can declare array and then a pointer to it. Array size must be enough to store the longest formula. Suppose, 100 will be enough:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char buf[100] = {0};
char *ptr = &buf[0];
scanf("%s", ptr);
// printf() gets a pointer as argument
printf("%s\n", ptr);
// printf() gets also a pointer as argument.
// When you pass arrays name without index to a function,
// you pass a pointer to array's first element.
printf("%s", buf);
return 0;
}
And as for rewriting letters in the end of the string. Here is a small program that does it. Pay attention at comments:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char buf[100] = {0};
char formula[100] = {0};
char compound[100] = {0};
char *ptr = &buf[0];
char *pFormula = &formula[0];
char *pCompound = &compound[0];
printf("Enter formula: ");
scanf("%s", pFormula);
printf("Enter chemical compound: ");
scanf("%s", pCompound);
// Copying the first chemical elements without the last
// several that will be replaced by another elements.
strncpy(ptr, pFormula, strlen(pFormula) - strlen(pCompound));
// Adding new compound to the first elements.
// Function also adds a null-terminated character to the end.
strncat(ptr, pCompound, strlen(pCompound));
printf("The new chemical compound is: ");
printf("%s", ptr);
return 0;
}
/* OUTPUT:
Enter formula: NaOH
Enter chemical compound: Cl
The new chemical compound is: NaCl
*/
In C, we use null-terminated strings. That is the "invisible", 0 value. Not ASCII "0", but the zero value, like 8-bit 0x00. You can represent it in literal text with '\0' or "\0" or unquoted 0, however, in a literal string it is redundant because most functions like strcmp() or strstr() or strcat() all expect and work with null terminated strings. Null char is the stops sign for the C standard string functions.
One easy way to implement this with C library calls is to test for existence of the substring and then test that substring's length, which verify it is at end of string.
Assume buf is some big string buffer, char buf[1024] and char *temp is a temporary variable.
temp = strstr(buf, "OH") returns the pointer to "OH" if exists in buf at any offset.
strlen(temp) Get length of temp, if at end of string, it will be 2 (doesn't include null terminator), so if the original string is "OHIO" or "SOHO" it wont match because it'll be 4 and 3 respectively.
The above is the core of the code, not the full robust implementation. You need to check for valid return values, etc.
char buf[1024];
char *temp;
strcpy(buf, "NaOH");
if((temp = strstr(buf, "OH")) != 0)
{
// At this point we know temp points to something that starts with "OH"
// Now see if it is at the end of the string
if(strlen(temp) == 2)
return true; // For C99 include stdbool.h
return false;
}
You could get obscure, and check for the null terminator directly, will be a smidge quicker. This code is safe as long as it is inside the if() for strstr(), otherwise never do this if you don't know a string is a least N characters long.
if(temp[2] == '\0')
return true; // For C99 include stdbool.h
As far as appending to a string, read the docs on strcat. Keep in mind with strcat, you must have enough space already in the buffer you are appending into. It isn't like C++ std::string or Java/C# string where those will dynamically resize as needed. In C, you get to do all of that yourself.