Issue sending c char* over USART - c

I am converting hex bytes into strings and sending them over uart to a terminal on my computer. I am receiving strange (almost correct) output and I cannot figure out why. This is all done in AVR. Here is the relevant code snippets:
while(true)
{
wan_usart_transmit_string(generateString());
}
Generate String:
char *generateString()
{
char *c = "";
char d[5];
sprintf(d, "%02X", 0x61);
c = strcat(c, d);
return c;
}
wan_usart_transmit_string:
void wan_usart_transmit_string(char * data)
{
unsigned char c = *data;
while (c)
{
while (!(UCSR0A & (1 << UDRE0)));
UDR0 = c;
c = *(++data);
}
}
Output: 61611611116111111116161161111611111111... etc...
The generateString method will eventually look ahead so many positions into a circular buffer and concatenate all of the ASCII values up to that position into a single string, which will subsequently be sent over USART. So it looks a little funky right now. This was just a test snippet to make sure it operates correctly before I made it function dynamically.

You need to allocate some memory for the string you are going to return from generateString().
Currently you are pointing c char* c = "" to a string literal and then you try to write into it causing undefined behavior.
Use malloc:
char *c = malloc(sizeof( *c )*10);
c[0] = '\0'; //nul terminate so you get an empty string
char d[5];
sprintf(d, "%02X", 0x61); //you can sprintf directly into c
c = strcat(c, d);
return c;

I was able to get the output I was looking for by using the following code in the generateString method:
char *generateString(){
char *c;
char d[5];
sprintf(d, "%02X", 0x61);
return c;
}

Related

C - String Reversal Function with Pointers Not Running

I understand that issues with string-reversal in C are numerous on this site, but I seem to be having a uniquely odd problem: the reversal logic simply does not run despite seeming to be set up correctly.
I wanted to write a program to reverse a given string in-place, by moving two pointers from the front and back of the string together:
void reverse_inplace_pointer(char *string_in, int length) {
char *back_ptr = string_in + (length - 1);
for (int i = 0; i < length / 2; i++) {
char swap_char = *back_ptr;
/* Dereference pointers to double-check values */
printf("\nCurrent front char: %c\nCurrent back char: %c", *string_in, *back_ptr);
/* Swap */
*back_ptr = *string_in;
*string_in = swap_char;
printf("\nSwap %u successful.", i + 1);
string_in++;
back_ptr--;
}
}
Calling this with a test string:
char *test_string = "TestMeTestMe";
nets the result:
Current front char: T
Current back char: e
And then execution just stops, at the assignment of the character pointed to by *string_in to the location pointed to by *back_ptr. If I remove the assignment, the program will quite happily move the pointers along the test string and print out each character without issue.
I looked for other examples of reassigning elements in arrays and found the logic in tutorials to be almost identical to mine. I copied this example wholesale as a test:
void steal_this_function(char *string) {
printf("\nEntering function:");
int length, c;
char *begin, *end, temp;
length = strlen(string);
begin = string;
end = string;
for (c = 0; c < (length - 1); c++) {
end++;
}
for (c = 0; c < length / 2; c++) {
temp = *end;
*end = *begin;
*begin = temp;
begin++;
end--;
}
}
The result is the same - it stops on the first iteration and the program exits. The only difference here is the reuse of the original pointer string, which is then walked to the end of the string with a for-loop.
Clearly I have some confusion here with pointers, but my understanding was that *back_ptr = *string_in; effectively says dereference the pointer string_in, then go to the location pointed to by back_ptr and store whatever it contains there.
I'm also compiling with the -Wall flag on GCC and it isn't finding any issues. This is frustrating because it should be very simple.
the function requires modifiable char array. It cannot be called when parameter is a pointer to thestring literal.
Some additional remarks.
This kind of function should return the reversed string. It is possible to use it as a parameter of another functions.
strlen returns size_t not int. Use the correct type in the loops as well
#include <string.h>
#include <stdio.h>
char *reverse(char *str)
{
char *end = str + strlen(str) - !!*str;
char *wrk = str;
while(end > wrk)
{
char tmp = *wrk;
*wrk++ = *end;
*end-- = tmp;
}
return str;
}
int main(void)
{
char str[] = "0123456789";
printf("%s\n", reverse(str));
}
Calling this with a test string: char *test_string = "TestMeTestMe";
That's the culprit, as noted in the comments, it declares a pointer to the memory where the string literal "TestMeTestMe" is stored, which you are not allowed to modify.
You should declare an array instead.
char test_string[] = "TestMeTestMe"; // Now test_string is a modifiable
// NULL-terminated char[13]
I'm also compiling with the -Wall flag on GCC and it isn't finding any issues.
-Wall it's not enough, but you can use -Wwrite-strings to catch this particular issue (see e.g. here).

What's wrong with my code function to make strcat function in C?

#include <stdio.h>
#include <stdlib.h>
char wordsum(char FW[256],char SW[256]){
int i;
int j=strlen(FW);
for (i=0;i<=strlen(SW);i++)
FW[i+j+1]=SW[i];
printf("%c",FW);
return FW;
}
int main()
{
char F[256];
char S[256];
printf("Enter the first word\n");
gets(F);
printf("Enter the Second word\n");
gets(S);
wordsum(F,S);
return 0;
}
I don't know what is wrong with my code to make strcat function. I hope to find the answer.
I assume that the code is written to learn more about the C language. If so, may I present an alternative implementation which does not use strlen(). The intention is to present some of the really nice features in the language. It may be a bit complicated to wrap ones head around the first time, but IIRC the code can be found in K&R's book The C Programming Language.
Here we go:
char* mystrcat(char *dest, const char *src)
{
char *ret = dest;
while (*dest)
dest++;
while ((*dest++ = *src++))
;
return ret;
}
The first while-loop finds the end of the destination string. The second while-loop appends the source string to the destination string. Finally, we return a pointer to the original dest buffer.
The function could've been even nicer if it didn't return a pointer.
void mystrcat(char *dest, const char *src)
{
while (*dest)
dest++;
while ((*dest++ = *src++))
;
}
HTH
There are several mistakes in your code. They are:
1) A function can't return an array in C and you don't need to do so. Change the return type from char to void of wordsum and erase the line return FW;
2) You want to print a string, right? Format specifier for string is %s. So write printf("%s",FW); instead of printf("%c",FW);.
3) Do this: FW[i+j]=SW[i];. Why did you add an extra 1 to i+j? Just think logically.
4) Add header file for strlen(), it's <string.h>.
5) Erase those asterisk marks before and after FW[i+j]=SW[i];.
There are a few problems in your function, I've changed and commented them below:
char *wordsum(char FW[256],char SW[256]){ // correct function type
int i;
int j=strlen(FW);
for (i = 0; i <= strlen(SW); i++)
FW[i+j] = SW[i]; //change 'i + j + 1' to 'i + j'
printf("%s",FW); //change format specifier as you are printing string not character
return FW;
}
Then dot forget to capture the returned pointer using a char* variable in the calling function (here main())
char *result;
result = wordsum(F,S);
printf("\n%s\n", result);
Working example: https://ideone.com/ERlFPE

Strcpy implementation in C

So, I have seen this strcpy implementation in C:
void strcpy1(char dest[], const char source[])
{
int i = 0;
while (1)
{
dest[i] = source[i];
if (dest[i] == '\0')
{
break;
}
i++;
}
}
Which to me, it even copies the \0 from source to destination.
And I have also seen this version:
// Move the assignment into the test
void strcpy2(char dest[], const char source[])
{
int i = 0;
while ((dest[i] = source[i]) != '\0')
{
i++;
}
}
Which to me, it will break when trying to assign \0 from source to dest.
What would be the correct option, copying \0 or not?
The code should look like as follows:
char * strcpy(char *strDest, const char *strSrc)
{
assert(strDest!=NULL && strSrc!=NULL);
char *temp = strDest;
while(*strDest++ = *strSrc++); // or while((*strDest++=*strSrc++) != '\0');
return temp;
}
You can NOT delete the second line char *temp = strDest; and directly return strDest. This will cause error for the returned content. For example, it will not return correct value (should be 22) will checking the length of returned char *.
char src_str[] = "C programming language";
char dst_str[100];
printf("dst_str: %d\n", strlen(strcpy(dst_str, src_str)));
Both copy the terminator, thus both are correct.
Note that strcpy2() does the assignment (the copying) first, then the comparison. So it will copy the terminator before realizing it did, and stopping.
Also, note that functions whose names start with str are reserved, so neither of these are actually valid as "user-level" code.
You're wrong. Both copy the \0 (NUL terminator) character. You have to copy the NUL terminator character always or your string will be broken: you'll never know when/where it ends.
Both copy the terminator, thus both are correct.
strcpy2() does the copying first, then the compares. Thus it will copy the terminator and stops.
The functions whose names start with str are reserved, so use any other variables or naming types
It is recommended not to advance the input pointers to the source and destination memory spaces, since the pointers will be used in main right away.
I've mentioned alternate methodical syntax, where in case someone might wonder the code output.
void strcpy1(char * s, char * p)
{
char * temp1 = s;
char * temp2 = p;
while(*temp1 != '\0')
{
*temp2 = *temp1;
temp1++;
temp2++;
}
*temp2 = '\0';
}
void main()
{
char * a = "Hello";
char b[10];
strcpy1(a,b);
printf("%s", b);
return 0;
}
Both strcpy1() and strcpy2() does the same. Both copy the NUL character to the end of the destination array.
Here is full implementation. You do not have to consider the \0 at the end in the first string, it will be copied automatically from the second string as per logic
//str copy function self made
char *strcpynew(char *d, char *s){
char *saved = d;
while ((*d++ = *s++) != '\0');
return saved; //returning starting address of s1
}
//default function that is run by C everytime
int main(){
//FOR STRCPY
char s1[] = "rahul"; //initializing strings
char s2[] = "arora"; //initializing strings
strcpynew(s1, s2);
printf("strcpy: %s\n", s1); //updated string after strcpy
}
You can use this code, the simpler the better !
Inside while() we copy char by char and moving pointer to the next. When the last char \0 will pass and copy while receive 0 and stop.
void StrCopy( char* _dst, const char* _src )
{
while((*_dst++ = *_src++));
}
char * strcpy(char *strDest, const char *strSrc)
{
assert(strDest!=NULL && strSrc!=NULL);
assert(strSrc + strlen(strSrc) < d || strSrc > strDest); // see note
char *temp = strDest;
while(*strDest++ = *strSrc++)
;
return temp;
}
// without the check on line 4, the new string overwrites the old including the null deliminator, causing the copy unable to stop.
Both copy the '\0'. That's what you have to do if you want to fully emulate the original strcpy

String (array) capacity via pointer

I am tring to create a sub-routine that inserts a string into another string. I want to check that the host string is going to have enough capacity to hold all the characters and if not return an error integer. This requires using something like sizeof but that can be called using a pointer. My code is below and I would be very gateful for any help.
#include<stdio.h>
#include<conio.h>
//#include "string.h"
int string_into_string(char* host_string, char* guest_string, int insertion_point);
int main(void) {
char string_one[21] = "Hello mother"; //12 characters
char string_two[21] = "dearest "; //8 characters
int c;
c = string_into_string(string_one, string_two, 6);
printf("Sub-routine string_into_string returned %d and creates the string: %s\n", c, string_one);
getch();
return 0;
}
int string_into_string(char* host_string, char* guest_string, int insertion_point) {
int i, starting_length_of_host_string;
//check host_string is long enough
if(strlen(host_string) + strlen(guest_string) >= sizeof(host_string) + 1) {
//host_string is too short
sprintf(host_string, "String too short(%d)!", sizeof(host_string));
return -1;
}
starting_length_of_host_string = strlen(host_string);
for(i = starting_length_of_host_string; i >= insertion_point; i--) { //make room
host_string[i + strlen(guest_string)] = host_string[i];
}
//i++;
//host_string[i] = '\0';
for(i = 1; i <= strlen(guest_string); i++) { //insert
host_string[i + insertion_point - 1] = guest_string[i - 1];
}
i = strlen(guest_string) + starting_length_of_host_string;
host_string[i] = '\0';
return strlen(host_string);
}
C does not allow you to pass arrays as function arguments, so all arrays of type T[N] decay to pointers of type T*. You must pass the size information manually. However, you can use sizeof at the call site to determine the size of an array:
int string_into_string(char * dst, size_t dstlen, char const * src, size_t srclen, size_t offset, size_t len);
char string_one[21] = "Hello mother";
char string_two[21] = "dearest ";
string_into_string(string_one, sizeof string_one, // gives 21
string_two, strlen(string_two), // gives 8
6, strlen(string_two));
If you are creating dynamic arrays with malloc, you have to store the size information somewhere separately anyway, so this idiom will still fit.
(Beware that sizeof(T[N]) == N * sizeof(T), and I've used the fact that sizeof(char) == 1 to simplify the code.)
This code needs a whole lot more error handling but should do what you need without needing any obscure loops. To speed it up, you could also pass the size of the source string as parameter, so the function does not need to calculate it in runtime.
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
signed int string_into_string (char* dest_buf,
int dest_size,
const char* source_str,
int insert_index)
{
int source_str_size;
char* dest_buf_backup;
if (insert_index >= dest_size) // sanity check of parameters
{
return -1;
}
// save data from the original buffer into temporary backup buffer
dest_buf_backup = malloc (dest_size - insert_index);
memcpy (dest_buf_backup,
&dest_buf[insert_index],
dest_size - insert_index);
source_str_size = strlen(source_str);
// copy new data into the destination buffer
strncpy (&dest_buf[insert_index],
source_str,
source_str_size);
// restore old data at the end
strcpy(&dest_buf[insert_index + source_str_size],
dest_buf_backup);
// delete temporary buffer
free(dest_buf_backup);
}
int main()
{
char string_one[21] = "Hello mother"; //12 characters
char string_two[21] = "dearest "; //8 characters
(void) string_into_string (string_one,
sizeof(string_one),
string_two,
6);
puts(string_one);
return 0;
}
I tried using a macro and changing string_into_string to include the requirement for a size argument, but I still strike out when I call the function from within another function. I tried using the following Macro:
#define STRING_INTO_STRING( a, b, c) (string_into_string2(a, sizeof(a), b, c))
The other function which causes failure is below. This fails because string has already become the pointer and therefore has size 4:
int string_replace(char* string, char* string_remove, char* string_add) {
int start_point;
int c;
start_point = string_find_and_remove(string, string_remove);
if(start_point < 0) {
printf("string not found: %s\n ABORTING!\n", string_remove);
while(1);
}
c = STRING_INTO_STRING(string, string_add, start_point);
return c;
}
Looks like this function will have to proceed at risk. looking at strcat it also proceeds at risk, in that it doesn't check that the string you are appending to is large enough to hold its intended contents (perhaps for the very same reason).
Thanks for everyone's help.

Using read() and write() system calls

So I have an assignment where I'm supposed to use read and write to read in lines from a file from stdin, sort it, then write it out to stdout with write. However, I can't get it to work.
I have to call a sort function on buf first to sort the string (which I got using read) before I re-output it. Can I treat buf as an array? Or does it not work like that? The reason is because I have to sort the string first.
int record_compare(const void *a, const void *b)
{
return (memcmp(a, b, num_bytes));
}
qsort(buf, num_elements, num_bytes, record_compare);
while (count < n - num_bytes)
{
i = memcmp(buf+count, buf+count + num_bytes, num_bytes);
if (i == 0)
count = count + num_bytes;
else
{
for (k = 0; k < num_bytes; k++)
{
printf("%c", buf[count]);
count++;
}
}
}
But since I got the string with read instead of something like fgets, can I still treat buf as an array? This is how the sort works on a normal array (it sorts it then prints out without repeats.
do {
c = read(0, buf+n, 1);
if (c != 0)
n++;
}
while (c != 0);
This is how I got buf.
The read() system call will read N bytes (1 byte at a time the way you've written the code). You will want to read up until you get a newline, or until you get zero bytes returned (which marks EOF), or until you get an error. Assuming you found a newline, you will start a new buffer.
You'll probably have an array of character pointers, and you'll probably allocate a new string for each line, and store the pointer in the array of character pointers.
enum { MAX_LINES = 1024 };
char *lines[MAX_LINES];
int n_lines = 0;
I'm assuming a fixed upper bound is OK; you could arrange to dynamically allocate the array of pointers too.
So, you'll need a function to read a line and store it.
Then you have a problem with your compare function. You'll be calling:
qsort(lines, n_lines, sizeof(char *), record_compare);
But record_compare() will be given two char ** values, so you need to use:
int record_compare(const void *v1, const void *v2)
{
const char *s1 = *(char **)v1;
const char *s2 = *(char **)v2;
return(strcmp(s1, s2));
}
Writing the sorted data is pretty straight forward. So, you need to concentrate on the line reader function.

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