I have an int value in the input and I want to save it to a buffer/array as a hexadecimal representation of ASCII.
I know how to print it in hex form but how can I save it like this to the buffer?
Here's the code I wrote:
int a = 98765;
char buffer[20];
m = sprintf(buffer, "%d", a);
printf("ASCII value in hex: ");
for(int i=0; buffer[i]!='\0'; i++)
{
printf("%02X", buffer[i]);
}
for eg if a=123 I want to get 313233
The first sprintf gave you an array of char, where each one is in the range 0x30 to 0x39. You want those to be expanded to the two characters '3' and '0', '3' and '1', etc.
You're expanding each char in the source to two chars in the output, which means you can't use the same destination buffer as source buffer -- you'd overwrite values in the source buffer before you processed them. You probably want to do the first sprintf into a separate array of char if you need the result in buffer. In your loop, change the printf to a sprintf to an index in buffer, incrementing your index by two each loop. (By two because that way you'll be overwriting the first iteration's string terminator with the second iteration's sprintf output, which will include a new string terminator. The sprintf in the last iteration of the loop will terminate the string for you.)
You didn't mention the longest input you're supposed to handle, or why you chose the array lengths you did. There's a danger of buffer overruns here if you accept inputs that are too long for your buffers (and don't forget to account for space for the null termination after the last character.)
You have already used sprintf to represent decimal values as a string; you can do the same for hexadecimal formatting.
Use the return value of sprintf() (which is the number of characters 'printed') to advance an index into the buffer as the destination for the next hex-digit pair, then iterate each character in the decimal buffer.
Note that, you can use the return value (m) from the first sprintf() for the iteration rather then testing for nul.
char hexbuffer[64];
int index = 0 ;
for(int i = 0; i < m; i++)
{
index += sprintf( &hexbuffer[index], "%02X", buffer[i] ) ;
}
Of course if the input includes only decimal digits (i.e. the user does not enter a negative value causing a - at buffer[0], you could simply insert a 3 ahead of each decimal digit, thus:
char hexbuffer[64];
for(int i = 0; i < m; i++)
{
bexbuffer[i * 2] = '3' ;
hexbuffer[i * 2 + 1] = buffer[i]
}
If only decimal digits were intended, you should change the type of a to unsigned and use the %u format specifier for the decimal input.
Related
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.....😊
How can I split character and variable in 1 line?
Example
INPUT
car1900food2900ram800
OUTPUT
car 1900
food 2900
ram 800
Code
char namax[25];
int hargax;
scanf ("%s%s",&namax,&hargax);
printf ("%s %s",namax,hargax);
If I use code like that, I need double enter or space for make output. How can I split without that?
You should be able to use code like this to read one name and number:
if (scanf("%24[a-zA-Z]%d", namax, &hargax) == 2)
…got name and number OK…
else
…some sort of problem to be reported and handled…
You would need to wrap that in a loop of some sort in order to get three pairs of values. Note that using &namax as an argument to scanf() is technically wrong. The %s, %c and %[…] (scan set) notations all expect a char * argument, but you are passing a char (*)[25] which is quite different. A fortuitous coincidence means you usually get away with the abuse, but it is still not correct and omitting the & is easy (and correct).
You can find details about scan sets etc in the POSIX specification of scanf().
You should consider reading a whole line of input with fgets() or POSIX
getline(), and then processing the resulting string with sscanf(). This makes error reporting and error recovery easier. See also How to use sscanf() in loops.
Since you are asking this question which is actually easy, I presume you are somewhat a beginner in C programming. So instead of trying to split the input itself during the input which seems to be a bit too complicated for someone who's new to C programming, I would suggest something simpler(not efficient when you take memory into account).
Just accept the entire input as a String. Then check the string internally to check for digits and alphabets. I have used ASCII values of them to check. If you find an alphabet followed by a digit, print out the part of string from the last such occurrence till the current point. And while printing this do the same with just a slight tweak with the extracted sub-part, i.e, instead of checking for number followed by letter, check for letter followed by digit, and at that point print as many number of spaces as needed.
just so that you know:
ASCII value of digits (0-9) => 48 to 57
ASCII value of uppercase alphabet (A-Z) => 65 to 90
ASCII value of lowercase alphabets (a-z)
=> 97 to 122
Here is the code:
#include<stdio.h>
#include<string.h>
int main() {
char s[100];
int i, len, j, k = 0, x;
printf("\nenter the string:");
scanf("%s",s);
len = strlen(s);
for(i = 0; i < len; i++){
if(((int)s[i]>=48)&&((int)s[i]<=57)) {
if((((int)s[i+1]>=65)&&((int)s[i+1]<=90))||(((int)s[i+1]>=97)&&((int)s[i+1]<=122))||(i==len-1)) {
for(j = k; j < i+1; j++) {
if(((int)s[j]>=48)&&((int)s[j]<=57)) {
if((((int)s[j-1]>=65)&&((int)s[j-1]<=90))||(((int)s[j-1]>=97)&&((int)s[j-1]<=122))) {
printf("\t");
}
}
printf("%c",s[j]);
}
printf("\n");
k = i + 1;
}
}
}
return(0);
}
the output:
enter the string: car1900food2900ram800
car 1900
food 2900
ram 800
In addition to using a character class to include the characters to read as a string, you can also use the character class to exclude digits which would allow you to scan forward in the string until the next digit is found, taking all characters as your name and then reading the digits as an integer. You can then determine the number of characters consumed so far using the "%n" format specifier and use the resulting number of characters to offset your next read within the line, e.g.
char namax[MAXNM],
*p = buf;
int hargax,
off = 0;
while (sscanf (p, "%24[^0-9]%d%n", namax, &hargax, &off) == 2) {
printf ("%-24s %d\n", namax, hargax);
p += off;
}
Note how the sscanf format string will read up to 24 character that are not digits as namax and then the integer that follows as hargax storing the number of characters consumed in off which is then applied to the pointer p to advance within the buffer in preparation for your next parse with sscanf.
Putting it altogether in a short example, you could do:
#include <stdio.h>
#define MAXNM 25
#define MAXC 1024
int main (void) {
char buf[MAXC] = "";
while (fgets (buf, MAXC, stdin)) {
char namax[MAXNM],
*p = buf;
int hargax,
off = 0;
while (sscanf (p, "%24[^0-9]%d%n", namax, &hargax, &off) == 2) {
printf ("%-24s %d\n", namax, hargax);
p += off;
}
}
}
Example Use/Output
$ echo "car1900food2900ram800" | ./bin/fgetssscanf
car 1900
food 2900
ram 800
I'm not familiar with C at all so this might be a simple problem to solve. I'm trying to take an input char* array of binary character sequences, ex. "0100100001101001", and output its relative string ("Hi"). The problem I'm having is coming up with a way to split the input into seperate strings of length 8 and then convert them individually to ulimately get the full output string.
char* binaryToString(char* b){
char binary[8];
for(int i=0; i<8; ++i){
binary[i] = b[i];
}
printf("%s", binary);
}
I'm aware of how to convert 8-bit into its character, I just need a way to split the input string in a way that will allow me to convert massive inputs of 8-bit binary characters.
Any help is appreciated... thanks!
From what I can tell, your binaryToString() function does not do what you'd want it to. The print statement just prints the first eight characters from the address pointed to by char* b.
Instead, you can convert the string of 8 bits to an integer, utilizing a standard C function strtol(). There's no need to convert any further, because binary, hex, decimal, etc, are all just representations of the same data! So once the string is converted to a long, you can use that value to represent an ASCII character.
Updating the implementation (as below), you can then leverage it to print a whole sequence.
#include <stdio.h>
#include <string.h>
void binaryToString(char* input, char* output){
char binary[9] = {0}; // initialize string to 0's
// copy 8 bits from input string
for (int i = 0; i < 8; i ++){
binary[i] = input[i];
}
*output = strtol(binary,NULL,2); // convert the byte to a long, using base 2
}
int main()
{
char inputStr[] = "01100001011100110110010001100110"; // "asdf" in ascii
char outputStr[20] = {0}; // initialize string to 0's
size_t iterations = strlen(inputStr) / 8; // get the # of bytes
// convert each byte into an ascii value
for (int i = 0; i < iterations; i++){
binaryToString(&inputStr[i*8], &outputStr[i]);
}
printf("%s", outputStr); // print the resulting string
return 0;
}
I compiled this and it seems to work fine. Of course, this can be done cleaner and safer, but this should help you get started.
I just need a way to split the input string in a way that will allow me to convert massive inputs of 8-bit binary characters.
You can use strncpy() to copy the sequence of '0' and '1' in a chunk of 8 characters at a time from the input string, something like this:
//get the size of input string
size_t len = strlen(b);
//Your input array of '0' and '1' and every sequence of 8 bytes represents a character
unsigned int num_chars = len/8;
//Take a temporary pointer and point it to input string
const char *tmp = b;
//Now copy the 8 chars from input string to buffer "binary"
for(int i=0; i<num_chars; ++i){
strncpy(binary, tmp+(i*8), 8);
//do your stuff with the 8 chars copied from input string to "binary" buffer
}
Maybe this can help. I didnt compile it but there is the idea. You can loop every 8 bit separately with while loop. And assign 8 bit to binary array with for loop. After that send this binary array to convert8BitToChar function to get letter equivalent of 8 bit. Then append the letter to result array. I'm not writing c for 3 year if there is mistakes sorry about that. Here pseudo code.
char* binaryToString(char* b){
char* result = malloc(sizeof(256*char));
char binary[8];
int nextLetter = 0;
while (b[nextLetter*8] != NULL) { // loop every 8 bit
for(int i=0; i<8; ++i){
binary[i] = b[nextLetter*8+i];
}
result[nextLetter] = 8bitToChar(binary));// convert 8bitToChar and append yo result
nextLetter++;
}
result[nextLetter] = '\0';
return result;
}
While writing a program i am filling the entries of a char array with digits. After doing so the length calculated for an array having no zero is correct but for an array starting with zero is zero!
Why is this result coming so!I am not able to interpret my mistake!?
int main()
{
int number_of_terms,no,j,i;
char arr[100];
char c;
scanf("%d",&number_of_terms);
for(i=0;i<number_of_terms;i++)
{
j=0;
while(c!='\n')
{
scanf("%d",&arr[j]);
if(c=getchar()=='\n')
break;
j++;
}
printf("Length is:%d\n",strlen(arr));
}
return 0;
}
for eg if i input my array elements as 4 5
lenght is 2
and if my array elements as 0 5
length is 0..
You are using "%d" in your format specifier, which produces an integer, and you are passing in the address of a character array. This is, exactly like your title says, undefined behaviour. In particular, the value zero will take up 4 of the cells in your string, and will write zero to all of those. Since the character with value zero is the end marker, you get zero length string. However, on another architecture, the second character would probably cause a crash...
If you want to store integers in an array, you should use int arr[...];. If you want to store characters, use "%c".
You are copying the value 0 into the array. This eqals the character '\0' which is used to terminate strings. What you want is to copy the character '0' (has the value 48, see an ascii table).
Change %d to %c to interpret the input has character instead of decimal.
scanf("%c",&arr[j]);
Also your "string" in arr is not zero terminated. After all the characters of your string, you have to end the string with the value 0 (here a decimal is correct). strlen needs it, because it determines the length of the string by traversing the array and counting up until it finds a 0.
I have a string that has ints and I'm trying to get all the ints into another array. When sscanf fails to find an int I want the loop to stop. So, I did the following:
int i;
int getout = 0;
for (i = 0; i < bsize && !getout; i++) {
if (!sscanf(startbuffer, "%d", &startarray[i])) {
getout = 1;
}
}
//startbuffer is a string, startarray is an int array.
This results in having all the elements of startarray to be the first char in startbuffer.
sscanf works fine but it doesn't move onto the next int it just stays at the first position.
Any idea what's wrong? Thanks.
The same string pointer is passed each time you call sscanf. If it were to "move" the input, it would have to move all the bytes of the string each time which would be slow for long strings. Furthermore, it would be moving the bytes that weren't scanned.
Instead, you need to implement this yourself by querying it for the number of bytes consumed and the number of values read. Use that information to adjust the pointers yourself.
int nums_now, bytes_now;
int bytes_consumed = 0, nums_read = 0;
while ( ( nums_now =
sscanf( string + bytes_consumed, "%d%n", arr + nums_read, & bytes_now )
) > 0 ) {
bytes_consumed += bytes_now;
nums_read += nums_now;
}
Convert the string to a stream, then you can use fscanf to get the integers.
Try this.
http://www.gnu.org/software/libc/manual/html_node/String-Streams.html
You are correct: sscanf indeed does not "move", because there is nothing to move. If you need to scan a bunch of ints, you can use strtol - it tells you how much it read, so you can feed the next pointer back to the function on the next iteration.
char str[] = "10 21 32 43 54";
char *p = str;
int i;
for (i = 0 ; i != 5 ; i++) {
int n = strtol(p, &p, 10);
printf("%d\n", n);
}
This is the correct behavior of sscanf. sscanf operates on a const char*, not an input stream from a file, so it will not store any information about what it has consumed.
As for the solution, you can use %n in the format string to obtain the number of characters that it has consumed so far (this is defined in C89 standard).
e.g. sscanf("This is a string", "%10s%10s%n", tok1, tok2, &numChar); numChar will contain the number of characters consumed so far. You can use this as an offset to continue scanning the string.
If the string only contains integers that doesn't exceed the maximum value of long type (or long long type), use strtol or strtoll. Beware that long type can be 32-bit or 64-bit, depending on the system.