C error: Exited with non-zero status - c

I know this code is quite simple; but that message won't stop appearing, can you help me find where the error is?
#include <stdio.h>
void Reverse(char * word[]) {
char temp = word[0];
for (int i = sizeof(word); i >= 0; i--){
for (int j = 0; j<= sizeof(word); j ++){
word[0] = word[i];
}
}
word[sizeof(word)] = temp;
printf("%s", word);
}
void main() {
Reverse(gets(stdin));
return 0;
}

gets returns char*. In this context - It is wrong to write char *[] in the function definition where you are supposedly passing a char array where input characters are being stored using gets. Also char *gets(char *str) - you need to pass a buffer to the gets where the inputted letters will be stored. You didn't pass one.
sizeof doesn't work here. It returns the size of a pointer (char*). You will have to use strlen() to get the length of the string inputted by gets.
More importantly - don't use gets - it's time to use something much safer than gets, namely fgets etc. Buffer overflow is not something you want to deal with.
Suppose you are passing an array of char* to the function reverse. Then the parameter would be char*[] which means nothing other than char** here. Here you will simply pass the char array which you will be using as buffer to gets.

Related

error: expected expression before ']' token when passing an array as an arguement in C

I am new to C. Was writing this so it takes the strings from the passed array and makes it a single sentence. But I got this error, I am not good with arrays in C. I can use some help from you guys. I did search an answer for this and couldn't find.
#include <stdint.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
char smash(char arr[20][20]) {
char tmp[sizeof(arr)/sizeof(arr[0])];
for (int i=0; i < sizeof(arr)/sizeof(arr[0]); i++) {
strcat(tmp, arr[i]);
strcat(tmp, " ");
}
return tmp;
}
int main(){
char list[][6] = {"hello", "world"};
printf("%s", smash(list[]));
}
Error
error: expected expression before ']' token
printf("%s", smash(list[]));
^
There are quite a number of errors in this small piece of code.
First, to address the compiler error: list[] is not a valid expression. If you want to pass list to the function, leave the braces out:
printf("%s", smash(list));
This will then bring up another error. The function is expecting a char [20][20] as it's argument, but that's not what you're passing in. Since arrays as parameters are converted to a pointer, the argument type is actually char (*)[20] i.e. a pointer to an array of char of size 20. Note also that this conversion only occurs for the outermost array dimension, not all.
Since you're passing in a char [2][6] which gets converted to a char (*)[6] this is a type mismatch. So change the parameter to char arr[][6].
Then you're attempting to get the size of the array parameter inside of the function:
sizeof(arr)/sizeof(arr[0])
Since arrays cannot be directly passed to a function due to the conversion mentioned earlier, arr is actually a pointer and not an array, so you won't get the result you expect from this. You'll need to pass the number of array elements as a separate parameter.
Then you're calling strcat on tmp. This function will only work if the destination already has a null terminated string in it. Since tmp was not initialized or written to prior to the first call to strcat, you end up reading uninitialized bytes and potentially past the end of the array which will trigger undefined behavior.
This can be fixed by setting the first byte of the array to 0 before the loop to make it an empty string:
tmp[0] = 0;
for ...
Then there's the problem with the return type. The function is declared to return a char but you're giving a char * to the return statement, and at the point the function is called it is passed to printf where the %s format specifier is expecting a char * parameter.
So change the return type of the function from char to char *.
Finally, you're returning a pointer to a local variable in the function. This variable's lifetime ends when the function returns, so the returned pointer is invalid and using it will also trigger undefined behavior.
You'll need change tmp to a pointer and dynamically allocate memory for it using malloc. This also means you'll need to save the return value of the function in a separate variable which you can then pass to printf to print and then pass to free to free the memory.
After making all this changes, the resulting code should look like this:
char *smash(char arr[][6], int len) {
// enough for len strings plus len spaces
char *tmp = malloc(sizeof(arr[0]) * len + len + 1);
tmp[0] = 0;
for (int i=0; i < len; i++) {
strcat(tmp, arr[i]);
strcat(tmp, " ");
}
return tmp;
}
int main(){
char list[][6] = {"hello", "world"};
char *result = smash(list, sizeof(list)/sizeof(list[0]));
printf("%s", result);
free(result);
return 0;
}

The below code runs encryption part well but when 'z' is entered it gives unexpected output and doesnt run decryption part

#include<stdio.h>
const char *encrypt(char *str);
const char *decrypt(char *str1);
int main()
{
char str[100],str1[100];
//Encryption
printf("Enter String for encryption\n");
gets(str);
encrypt(str);
printf("%s after encryption is %s\n",str,encrypt(str));
//Encryption
printf("Enter String for decryption\n");
gets(str1);
decrypt(str1);
printf("%s after decryption is %s",str1,decrypt(str1));
return 0;
}
const char *encrypt(char *str)
{
char en[100];
int i=0;
for(;i<100;i++)
{
en[i]=str[i]+1;
}
en[i]='\0';
return en;
}
const char *decrypt(char *str1)
{
char de[100];
int i=0;
for(;i<100;i++)
{
de[i]=str1[i]-3;
}
de[i]='\0';
return de;
}
You are returning a pointer to automatic variables en and de which are stored in the stack. This in turn means after returning from the functions encrypt and decrypt their place in the memory can be used by any other variable.
so to correct that, you need to define en and de as static.
const char *encrypt(char *str){
static char en[100];
int i=0;
for(;i<100;i++){
en[i]=str[i]+1;
}
en[i]='\0';
return en;
}
const char *decrypt(char *str1){
static char de[100];
int i=0;
for(;i<100;i++){
de[i]=str1[i]-3;
}
de[i]='\0';
return de;
}
Though a more suitable and safer way to implement that would be:
#include<stdio.h>
void encrypt(char *str, char *encStr);
void decrypt(char *str1, char* decStr);
int main()
{
char str[100], encDecStr[100];
//Encryption
printf("Enter String for encryption\n");
scanf("%s", str);
encrypt(str, encDecStr);
printf("%s after encryption is %s\n",str,encDecStr);
//Decryption
printf("Enter String for decryption\n");
scanf("%s", str);
decrypt(str, encDecStr);
printf("%s after decryption is %s",str,encDecStr);
return 0;
}
void encrypt(char *str, char *encStr)
{
for(char *c = str; *c != '\0'; c++)
{
*encStr++ = *c + 1;
}
*encStr='\0';
}
void decrypt(char *str1, char* decStr)
{
for(char *c = str1; *c != '\0'; c++)
{
*decStr++ = *c - 1;
}
*decStr++='\0';
}
Note: The code was not fully tested for different use cases.
There are quite a number of errors in your code:
Returning arrays with local storage duration:
The array's life time ends (i.e. it ceases to exist) as soon as you exit from the function, thus the pointer returned is dangling, reading from it is undefined behaviour
You write beyond the bounds of your local array: en[i] = '\0' with i being 100 after the loop is out of the range of valid indices from 0 to 99, which again is undefined behaviour.
You have differing offsets for encrypting (1) and decrypting (3).
Simply adding an offset without further checks (or modulo operations) will produce different character sets for input and output strings (might be your least problem...).
You always en-/decode the entire array, which is more than actually needed. Additionally the terminating null character then is encoded as well, resulting in different lengths of input and output and garbage at the end of encoded string.
Use of gets is dangerous as it allows a user to write beyond the input array, resulting in undefined behaviour. This is the reason why it actually has been removed from C language with C11 standard – which introduces a safe alternative gets_s. Yet another alternative (especially for pre-C11 code) is fgets.
For the dangling pointer problem there are several options:
Making the array static (as mentioned already):The disadvantage of this approach is that the function is not thread-safe any more. Additionally calling the function more than once overwrites previous results, if you haven't evaluated already or copied them they are lost.
Returning a dynamically allocated array, see malloc function. This comes with the risk of the caller forgetting to free the allocated memory again, which would result in a memory leak
Changing the input array in place: Disadvantage of is having to copy the input into a backup if it is yet needed afterwards.
Letting the caller provide the buffer.
Last option is most flexible and most idiomatic one, so I'll concentrate on this one:
void caesar(char const* input, char* output, int offset)
{
int const NumChars = 'z' - 'a';
offset = offset % NumChars + NumChars;
// ^ assures valid range, can still be negative
// ^ assures positive value, possibly
// out of range, but will be fixed later
for(;*input; ++input, ++output)
{
int c = *input - 'a';
// distance from character a
c = (c + offset) % NumChars;
// ^ fixes too large offset
*output = 'a' + c;
}
// as now iterating *until* null character found we now *need*
// to add it (in contrast to original version with whole array!)
*output = 0;
}
This variant includes an interesting idea: Let the caller define the offset to be applied! The modulo operation assures the valid range of the character, no matter how large the offset is. The great thing about: If a user encoded with some number n, exactly the same function can be used to decode again with -n (which is why I simply named it caesar according to the algorithm it implements). Note that this is untested code; if you find a bug please fix yourself...
Be aware, though, that this code still has two weaknesses:
It assumes ASCII encoding or compatible – at least one where letters a to z are in contiguous range, a is first character, z is last one. This is not true for all encodings, though, consider the (in-?)famous EBCDIC.
It assumes all input is in range of Latin minuscules only (from a - z), it does not consider white-space, digits, capital letters, punctuation marks... You might want to include special handling for all of these or at least the ones you might use.
You could fix these e.g. (many other variants are thinkable as well) by
converting all characters to either upper or lower case (toupper((unsigned char)*input) – assuming case doesn't matter)
search in an array of valid characters ("ABC...XYZ012...89") for the appropriate index and if found encode like above with NumChars being array length, otherwise (whitespace, punctuation) just leave as is.
In any case, the function would then be called like:
char input[128]; // prefer powers of 2 for array lengths...
char en[sizeof(input)];
char de[sizeof(input)];
gets_s(input, sizeof(input));
caesar(input, en, 7);
// en contains encrypted string, could e.g. be printed now
caesar(en, de, -7);
// de contains decrypted string
// you could even encode/decode in place:
caesar(input, input, 7);
// just be aware that this will fail, though, for the case if input
// and output overlap and input starts before output, as then at
// some point already encoded values will be read as input again:
// caesar(input, input + 1, 7) fails!!!
There's some issues in your code :
Not a very big issue for a beginner , but you should avoid gets function.
Because it doesn't check the input , it can cause buffers overflow and various security problems , try using fgets instead.
In encrypt , and decrypt functions , you are returning the address of an array located in the stack of the function , look :
const char *encrypt(char *str){
char en[100];
int i=0;
for(;i<100;i++){
en[i]=str[i]+1;
}
en[i]='\0';
return en;
}
Here , Since the en array is declared inside the function , after the return you may get garbage string when trying to read it.
The solution here , is to either malloc it , or declare a static array outside the function , and initialize it.
You are encrypting by adding 1 to the value of the string , and decrypt it by retrieving 3 . I don't know if this is what you intended to do.
Here's a new version of your code , try to check if it suits your need :
#include <stdio.h>
#include <stdlib.h>
#include <memory.h>
#include <string.h>
static char de[100] , en[100] ;
const char *decrypt(char *str1){
memset(de , 0 , 100) ;
int i=0;
for(;i<strlen(str1);i++){
de[i]=str1[i]-1;
}
de[i]='\0';
return (const char*) de;
}
const char* encrypt(char* str){
memset(en , 0 , 100) ;
int i=0;
for(;i<strlen(str);i++){
en[i]=str[i]+1;
}
en[i]='\0';
return (const char*) en;
}
int main(){
char str[100],str1[100];
//Encryption
printf("Enter String for encryption\n");
gets(str);
encrypt(str);
printf("%s after encryption is %s\n",str,encrypt(str));
//Encryption
printf("Enter String for decryption\n");
gets(str1);
decrypt(str1);
printf("%s after decryption is %s",str1,decrypt(str1));
return 0;
}
Your code does not handle a special case for the character 'z'. Thus
en[i]=str[i]+1;
Causes the character '{' to be written to the array en instead. For learning more about why this happens, I recommend you look at ASCII tables and looking at the integer values for alphabets.
Secondly, did you mean to type -3 in there?
de[i]=str1[i]-3;
This won't work if you're planning on using the decrypt() function to decrypt strings that you made using encrypt() because you're adding 1 to the character while encrypting and then subtracting a different number when decrypting, so the result will appear garbled.
I rewrote your code for you, since this is a beginner program, I made as little changes as possible so you can understand it. I WOULD HIGHLY RECOMMEND NOT USING gets() though... See here.
#include<stdio.h>
const char *encrypt(char *str);
const char *decrypt(char *str1);
int main()
{
char str[100],str1[100];
//Encryption
printf("Enter String for encryption\n");
gets(str); // DON'T USE GETS!!! USE fgets(str, 100, stdin);
encrypt(str);
printf("%s after encryption is %s\n", str, encrypt(str));
//Encryption
printf("Enter String for decryption\n");
gets(str1); // DON'T USE GETS!!! USE fgets(str, 100, stdin);
decrypt(str1);
printf("%s after decryption is %s", str1, decrypt(str1));
return 0;
}
const char *encrypt(char *str)
{
char en[100];
int i=0;
for(; i<100; i++)
{
if (str[i] == 'z')
{
en[i] = 'a';
continue;
}
en[i] = str[i] + 1;
}
en[i] = '\0';
return en;
}
const char *decrypt(char *str1)
{
char de[100];
int i=0;
for(; i<100; i++)
{
if (str[i] == 'a')
{
en[i] = 'z';
continue;
}
de[i] = str1[i] - 1;
}
de[i] = '\0';
return de;
}
Some criticisms
Like I said, gets() is really bad... See here for more details. Although it might be too complicated for you... A better alternative is fgets!
fgets(str, num, stdin)
takes user input from the console, and then stores it inside the array str, which must be large enough to store at least num characters. Don't worry about stdin if you don't know what that means. But be sure to always write it when using fgets as an alternative to gets
Like others have already posted, albeit using more technical jargon, it's a bad idea to declare an array inside a function and then return that array. You know the function ends when the return statement is hit, and at that point all the variables that were declared inside the function will get destroyed.
That doesn't necessarily mean that you can't read the data that was in them, but it becomes a probabilistic game where there's a teeny-tiny chance that the array will get corrupted after the function exits and before the data in that array is read. This is technically Undefined Behaviour.
I hope you know about pointers. You can modify the array which you passed as a parameter directly and then return that array, thus avoiding accessing an array outside it's lifetime.

Program to conjugate

I am trying to write a program that will conjugate a verb in multiple forms.
So I write a function that will allow me to get the part that is kepted and conjugate it. Fixed rule : whole word except last 2 chars.
I am used to OO, and I can't seem to make it work, while it seems a basic program.
I obtain something with weird : here is a screen at the end of the execution, that will be more explicit
I think I missed a little something in my course (probably in the char[] part...), that has a huge impact, but I can't seem to find it.
I am opened to all observations on my code, since I am beginning, and I prefer going on a solid basis right now, better that later.
Here is the code
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
void RacineVerbe(char verbeEntier[], char dest[]);
int myStrLen(char *s);
int main()
{
char *string;
char *racine;
string = (char*)malloc(200*sizeof(char));
racine = (char*)malloc(200*sizeof(char));
printf("Quel verbe?\n");
scanf("%s", string);
RacineVerbe(string, racine);
printf("%s", racine);
printf("%sASSE\n", racine);
printf("%sASSES\n", racine);
printf("%sAT\n", racine);
printf("%sASSIONS\n", racine);
printf("%sASSIEZ\n", racine);
printf("%sASSENT\n", racine);
return 0;
}
void RacineVerbe(char verbeEntier[], char dest[]){
int i;
int l = myStrLen(verbeEntier);
for( i = 0; i < l -2 ; i++){
dest[i] = verbeEntier[i];
}
dest[i+1] = "\0";
}
int myStrLen(char *s){
int i = 0;
while(*s++)i++;
return i;
}
The problem is in you RacineVerbe because you assigned string literal to single character (not a poitner to characters).
In that situation, string literal returns you the address where it is in memory and you assigned to dest[i+i] LSB byte of that address and it may be or may not be visible character. I can assure, your compiler gave you at least warning for that.
Second problem is where you did assignment. You should do it to dest[i] as i was last time incremented in for loop before check failed and therefore i already points to place where 0 should be written.
void RacineVerbe(char verbeEntier[], char dest[]){
int i;
int l = myStrLen(verbeEntier);
for( i = 0; i < l -2 ; i++){
dest[i] = verbeEntier[i];
}
dest[i] = 0; //This line was rewritten.
}
And as already mentioned, try to NOT cast return result of malloc.
You invoked UB when you wrote dest[i+1] = "\0"; since dest[i+1] expects a char and you assigned "\0" into it, which is a string literal. Replace it with '\0'
Note that string = (char*)malloc(200*sizeof(char)); ---> string = malloc(200); since casting is not needed in malloc in C, and considered bad (as you can see here) and also sizeof(char) is, by definition, 1

Function that prints reverse of a string/char array in C

I am rather new to the C language right now and I am trying some practice on my own to help me understand how C works. The only other language I know proficiently is Java. Here is my code below:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
const char * reverse(char word[]);
const char * reverse(char word[]) {
char reverse[sizeof(word)];
int i, j;
for (i = sizeof(word - 1); i <= 0; i--) {
for (j = 0; j > sizeof(word - 1); j++) {
reverse[i] = word[j];
}
}
return reverse;
}
int main() {
char word[100];
printf("Enter a word: ");
scanf("%s", word);
printf("%s backwards is %s\n", word, reverse(word));
return 0;
}
When the user enters a word, the program successfully prints it out when i store it but when i call the reverse function I made it doesnt return anything. It says on my editor the address of the memory stack is being returned instead and not the string of the array I am trying to create the reverse of in my function. Can anyone offer an explanation please :(
sizeof(word) is incorrect. When the word array is passed to a function, it is passed as a pointer to the first char, so you are taking the size of the pointer (presumably 4 or 8, on 32- or 64-bit machines). Confirm by printing out the size. You need to use strlen to get the length of a string.
There are other problems with the code. For instance, you shouldn't need a nested loop to reverse a string. And sizeof(word-1) is even worse than sizeof(word). And a loop that does i-- but compares i<=0 is doomed: i will just keep getting more negative.
There are multiple problems with your reverse function. C is very different from Java. It is a lot simpler and has less features.
Sizes of arrays and strings don't propagate through parameters like you think. Your sizeof will return wrong values.
reverse is an identifier that is used twice (as function name and local variable).
You cannot return variables that are allocated on stack, because this part of stack might be destroyed after the function call returns.
You don't need two nested loops to reverse a string and the logic is also wrong.
What you probably look for is the function strlen that is available in header string.h. It will tell you the length of a string. If you want to solve it your way, you will need to know how to allocate memory for a string (and how to free it).
If you want a function that reverses strings, you can operate directly on the parameter word. It is already allocated outside the reverse function, so it will not vanish.
If you just want to output the string backwards without really reversing it, you can also output char after char from the end of the string to start by iterating from strlen(word) - 1 to 0.
Edit: Changed my reverse() function to avoid pointer arithmetic and to allow reuse of word.
Don't return const values from a function; the return value cannot be assigned to, so const doesn't make sense. Caveat: due to differences between the C and C++ type system, you should return strings as const char * if you want the code to also compile as C++.
Arrays passed as params always "decay" to a pointer.
You can't return a function-local variable, unless you allocate it on the heap using malloc(). So we need to create it in main() and pass it as a param.
Since the args are pointers, with no size info, we need to tell the function the size of the array/string: sizeof won't work.
To be a valid C string, a pointer to or array of char must end with the string termination character \0.
Must put maximum length in scanf format specifier (%99s instead of plain %s — leave one space for the string termination character \0), otherwise vulnerable to buffer overflow.
#include <stdio.h> // size_t, scanf(), printf()
#include <string.h> // strlen()
// 1. // 2. // 3. // 4.
char *reverse(char *word, char *reversed_word, size_t size);
char *reverse(char *word, char *reversed_word, size_t size)
{
size_t index = 0;
reversed_word[size] = '\0'; // 5.
while (size-- > index) {
const char temp = word[index];
reversed_word[index++] = word[size];
reversed_word[size] = temp;
}
return reversed_word;
}
int main() {
char word[100];
size_t size = 0;
printf("Enter a word: ");
scanf("%99s", word); // 6.
size = strlen(word);
printf("%s backwards is %s\n", word, reverse(word, word, size));
return 0;
}

building a string out of a variable amount of arguments

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdarg.h>
int main(int argc, char * argv[])
{
char *arr[] = { "ab", "cd", "ef" };
char **ptr, **p, *str;
int num = 3;
int size = 0;
ptr = calloc(num, 4);
p = ptr;
for (; num > 0; num--)
size += strlen(*(p++) = arr[num - 1]);
str = calloc(1, ++size);
sprintf(str, "%s%s%s", ptr[0], ptr[1], ptr[2]);
printf("%s\n", str);
return 0;
}
output: "efcdab" as expected.
now, this is all fine and suitable if the argument count to sprintf is predetermined and known. what i'm trying to achieve, however, is an elegant way of building a string if the argument count is variable (ptr[any]).
first problem: 2nd argument that is required to be passed to sprintf is const char *format.
second: the 3rd argument is the actual amount of passed on arguments in order to build the string based on the provided format.
how can i achieve something of the following:
sprintf(str, "...", ...)
basically, what if the function receives 4 (or more) char pointers out of which i want to build a whole string (currently, within the code provided above, there's only 3). that would mean, that the 2nd argument must be (at least) in the form of "%s%s%s%s", followed by an argument list of ptr[0], ptr[1], ptr[2], ptr[3].
how can make such a 'combined' call, to sprintf (or vsprintf), in the first place? things would be easier, if i could just provide a whole pointer array (**ptr) as the 3rd argument, instead.. but that does not seem to be feasible? at least, not in a way that sprintf would understand it, so it seems.. as it would need some special form of format.
ideas / suggestions?
karlphillip's suggestion of strcat does seem to be the solution here. Or rather, you'd more likely want to use something like strncat (though if you're working with a C library that supports it, I'd recommend strlcat, which, in my opinion, is much better than strncat).
So, rather than sprintf(str, "%s%s%s", ptr[0], ptr[1], ptr[2]);, you could do something like this:
int i;
for (i = 0; i < any; i++)
strncat(str, arr[i], size - strlen(str) - 1);
(Or strlcat(str, arr[i], size);; the nice thing about strlcat is that its return value will indicate how many bytes are needed for reallocation if the destination buffer is too small, but it's not a standard C function and a lot of systems don't support it.)
There's no other way to do this in C without manipulating buffers.
You could, however, switch to C++ and use the fabulous std::string to make your life easier.
Your first problem is handled by: const char * is for the function, not you. Put together your own string -- that signature just means that the function won't change it.
Your second problem is handled by: pass in your own va_list. How do you get it? Make your own varargs function:
char *assemble_strings(int count, ...)
{
va_list data_list;
va_list len_list;
int size;
char *arg;
char *formatstr;
char *str;
int i;
va_start(len_list, count);
for (i = 0, size = 0; i < count; i++)
{
arg = va_arg(len_list, char *);
size += strlen(arg);
}
va_end(len_list);
formatstr = malloc(2*count + 1);
formatstr[2*count] = 0;
for (i = 0; i < count; i++)
{
formatstr[2*i] = '%';
formatstr[2*i+1] = 's';
}
str = malloc(size + 1);
va_start(data_list, count);
vsprintf(str, formatstr, data_list);
va_end(data_list);
free(formatstr);
return(str);
}
You'll need some way to terminate the varargs, of course, and it's much easier to just pass it to vsprintf if the string list is entirely within the varargs -- since standard C requires at least one regular argument.
The loop I would use for the final copy into str would be something like:
for(i=0, p=str; i < num; i++)
p += sprintf(p, "%s", ptr[i]);
or
for(i=0, p=str; i < num; i++)
p += strlen(strcpy(p, ptr[i]));
rather than trying to deal with a variable number of arguments in a single call to sprintf.

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