I'm passing values beteween childs and need to store some values to later use.
the definitions and use in functions
char fouts[MAX_SIZE][10];
the function where i give the array the values:
void connect(char *nodo, char *out[], int nouts) {
(...)
for(i=0;i<nouts;i++) {
fouts[fnum][i] = out[i];
}
(...)
and the function where i'm trying to use them:
void disconnect(char *nodo, char *remover){
char *outs[10];
nouts = fnouts[getfnum];
int m =0;
for(i=0;i<nouts;i++) {
if(strcmp(fouts[getfnum][i],nodo) != 0) { outs[m] = fouts[getfnum][i]; m++ ; }
}
no matter what i did to try to fixm everytime it tries to execute this last for, it gives a segmentation fault.
have tried somethings (read fouts[getfnum][0] for example directly and gives a segmentaton fault, but fouts[getfnum] gives "trash")
check the value after it been atributed fouts[fnum][i] = out[i]; here and it checks out, so i guess that part is ok).
don't know if its something obvious or not, but any help?
You are mixing char and char*.
fouts[fnum][i] is a char
and
out[i] is a char pointer
So in this line
fouts[fnum][i] = out[i];
you assign a char pointer to a char which is illegal.
And in this line
if(strcmp(fouts[getfnum][i],nodo) != 0)
you pass a char (i.e. fouts[getfnum][i]) to strcmp.
That is not legal as strcmp expects a char*
From the posted code, it is hard to tell how to fix the problems. Maybe you just need:
char* fouts[MAX_SIZE][10];
Related
I'm trying to read a string from argv. It should be something like a mathematical function like "-x^2-5" etc.
I've got something like this:
void test(char *function){
int temp = 0;
for (int i = 0; i < strlen(function); i++){
if((function[i] != '+') && (function[i] != ...){
// function[i] is a digit, but still as a char
temp = atoi(function[i]);
}
}
//...
}
int main(int argc, char **argv){
test(argv[1]);
//...
}
This works quite well until the last lap of the loop.
If I use printf("%c", function[i]);, it sais, it's 5.
But atoi(function[i]) gives a segfault. Why?
Right. Let's first take a look at the signature of atoi.
int atoi(const char *str);
Now, what you are passing to it is function[i], and since function is of type char *, function[i] is a char. And your character is most likely not a proper character pointer.
What you'd want to pass to atoi is instead a const char *, which you can get by calling atoi(&function[i]);
As already said by #sham1, atoi expects its parameter to be a C string, this is a null terminated char array, while you have a single char.
But digits are special, because they are required to have consecutive codes. So if you know that a character (say c) is a digit, its value is c - '0'. So you can write:
if((function[i] != '+') && (function[i] != ...){
// function[i] is a digit, but still as a char
temp = function[i] - '0'; // get the int value of a digit character
}
Besides the bug pointed out in other answers, atoi doesn't have any reliable error handling, which is why it is one of the standard library functions that should never be used.
Just forget that you ever heard about atoi and replace it with:
char* endptr;
int result = strtol(str,&endptr,10);
if(endptr == str)
{
// some conversion error
}
strtol is also able to convert an initial part of the string if it contains valid numbers.
When I try to get a string from a function which has type void*, I only get the first character.
I tried to copy string manually without using strcpy() it gives me same problem
struct arr {
int first
int last.
void **val;
};
//I have a function which is called
void *inspect_arr(const arr *ar, int position)
{
....
return ar->val[offset];
}
//I want to inspect the array so that I can compare the strings
int main()
{
char *str = calloc(10,sizeof(char));
char *k = *(char*)inspect_arr(...) //I have a string in the array
// strcpy(str,k); Doesn't work
strcmp(str,k); Invalid read from valgrind
//If array has an integer type then I would write my code like this:
//int a = *(int*)inspect_arr(...) but I can not do the same thing for char
}
I get Segmentation fault when I run the program.
One problem is that first * in *(char*)inspect_arr(...).
There you dereference the pointer, getting only the first character. It's equivalent to ((char*)inspect_arr(...))[0].
To make it more troublesome, you then assign this single character to the pointer k. This doesn't make k point to the character, it make k point to whatever address corresponds to that characters encoded value. With e.g. ASCII and the character 'd' the initialization is equivalent to
char *k = 100;
Good evening, I have 2 functions and each of them accepts as argument a pointer to char:
char pointer[255];
func1(char* pointer)
{
...
memcpy(pointer,some_char,strlen(something));
return;
}
func2(char* pointer)
{
...
if (pointer==someother_char) exit(0); //FAILs
//also I have
if(pointer==someother_pointer2char); // FAILs
}
Now I've tried strstr,strcmp etc... doesn't work. Wanted to try memcmp but I don't have static len. As I have to compare char* to char and char* to char* I would be needing two solutions right?
So, how to compare these pointers (actually pointees) in shortest possible way?
Thanks.
E D I T
Thanks to wallacer and Code Monkey now for char* to char comparison I use following:
func1(char* ptr){
char someother_char[255];
char *ptr_char = NULL; //I have to strcmp a few values so this is why I initialize it first
...
ptr_char = someother_char;
if (strcmp(ptr,ptr_char) == 0) //gtfo and it does...
...
ptr_char = some2nd;
if(strcmp...
Any suggestions maybe... (hmm external function for comparing?)
Suggestion1(by Code Monkey)
#include <stdio.h>
int main(void) {
char tempchar[255];
tempchar[0] = 'a';
tempchar[1] = 'b';
tempchar[2] = '\0';
char *ptr_char;
ptr_char = &tempchar[0];
printf("%s", ptr_char);
return 0;
}
You need to use strcmp. Not seeing how you tried to use it, this is how you should use it:
char *someother_char = "a";
char *pointer = "a";
if (strcmp(pointer, someother_char) == 0) { // match!
}
else { // not matched
}
to then do the comparison with a char, you have to promote to a char*:
char *someother_char1;
char test = 'a';
char *pointer = "a";
strncpy((char*)test,someother_char1,sizeof(test));
if (strcmp(pointer, someother_char1) == 0) { // match!
}
else { // not matched
}
if you want to use the char array then you have to de-reference:
char char_array[255];
// don't forget to fill your array
// and add a null-terminating char somewhere, such as char_array[255] = '\0';
char *ptr_somechar = &char_array[0];
char *pointer = "a";
if (strcmp(pointer, ptr_somechar) == 0) { // match!
} else { // not matched
}
Well right off the bat, if you want to compare the pointees, you need to dereference them. This means to compare the actual char value, you'll have to call
if (*pointer == someother_char)
However this will only compare the first char in the array, which is probably not what you want to do.
To compare the whole thing strcmp should work
char* someother_str = "hello strstr";
if(strcmp(pointer, someother_str) == 0) {
// do something
}
Make sure your other string is declared as a char*
More info: http://www.cplusplus.com/reference/clibrary/cstring/strcmp/
Edit: as per your comment. comparing char* and char doesn't really make sense. One is a character value, the other is an address in memory. Do do so, you can either dereference the char* or reference the value variable.
char c;
char* ptr;
// dereference ptr
if ( c == *ptr ) {
...
}
// reference the value
if ( &c == ptr ) {
}
The first method checks if the values are the same. The second checks if ptr is in fact pointing to the memory containing c ie. is ptr a pointer to c
Hope that helps
Use function srtncmp no srtcmp.
int res = strncmp(str, "¿Cuál es tu nombre? ", 100);
See the next link
compare strings
Strings are null terminated. When you use such kind of strings, it's not a good idea to mixing with other memory copy functions.
Once you do the memcpy operation, please note that your destination string will not be null terminated.
memcmp is a fast operations. Otherwise yo can simply loop through each character and quit upon finding a difference.
To use strcmp, please make sure that both the strings are null terminated. Otherwise it will lead to some crash.
I suggest you to use string functions like strcmp,strlen, strcpy to deal with strings because for that it's actually implemented.
You can't compare two pointers unless both pointers are referring to same memory location. Pointer is just a address to a memory location. What you really want to do is that, to compare the contents rather than compare the address where it's stored. So please use strcmp but again I warn you make sure that it's null terminated.
I have made strstr() function but the program does not give any output,just a blank screen.Please have a look at the code.
#include<stdio.h>
#include<conio.h>
const char* mystrstr(const char *str1, const char *str2);
int main()
{
const char *str1="chal bhai nikal";
const char *str2="nikal",*result;
result=mystrstr(str1,str2);
printf("found at %d location",(int*)*result);
getch();
return 0;
}
const char * mystrstr(const char *s1, const char *s2)
{
int i,j,k,len2,count=0;
char *p;
for(len2=0;*s2!='\0';len2++);//len2 becomes the length of s2
for(i=0,count=0;*s1!='\0';i++)
{
if(*(s1+i)==*s2)
{
for(j=i,k=0;*s2!='\0';j++,k++)
{
if(*(s1+j)==*(s2+i))
count++;
if(count==len2)
{
p=(char*)malloc(sizeof(char*));
*p='i';
return p;
}
}
}
}
return NULL;
}
The line with this comment:
//len2 becomes the length of s2
is broken. You repeatedly check the first character of s2. Instead of *s2, try s2[len2].
Edit: as others have said, there are apparently a lot more things wrong with this implementation. If you want the naive, brute-force strstr algorithm, here's a concise and fast version of it:
char *naive_strstr(const char *h, const char *n)
{
size_t i;
for (i=0; n[i] && h[i]; i++)
for (; n[i] != h[i]; h++) i=0;
return n[i] ? 0 : (char *)h;
}
It looks like this is an exercise you're doing to learn more about algorithms and C strings and pointers, so I won't solve those issues for you, but here are some starting points:
You have an infinite loop when calculating len2 (your loop condition is *s2 but you're never changing s2)
You have a similar issue with the second for loop, although I you have an early return so it might not be infinite, but I doubt the condition is correct
Given you want to behave like strstr(), you should return a pointer to the first string, not a new pointer you allocate. There is no reason for you to allocate during a function like strstr.
In main() if you want to calculate the position of the found string, you want to print result-str1 unless result is NULL). (int*)*result makes no sense - result should be a pointer to the string (or NULL)
You also need to change this line:
if(*(s1+j)==*(s2+i))
to this:
if(*(s1+j)==*(s2+k))
As already mentioned, the return value is a bit odd. You are returning a char* but kind of trying to put an integer value in it. The result doesn't make sense logically. You should either return a pointer to the location where it is found (no malloc needed) or return the integer position (i). But returning the integer position is not the "typical" strstr implementation.
I have a doubt in my program
#include<stdio.h>
int myFunc(char **);
main()
{
char *a[2]={"Do","While"};
myFunc(a);
}
int myFunc(char **P)
{
/* Here I want to print the strings passed but I'm unable to
print the strings I just tried the below statement which
printed just the first letter which is 'D'*/
printf("%c",**P);
}
when i tried
printf("%s",**P);
I am getting run time error. so can anyone please help me out?
Thanks
Madhu
Put size as parameter to allow the function to know how many strings you have in your array. Then, you should iterate the array and print each one.
int myFunc( char** p, int size)
{
for( int i = 0; i < size; ++i)
{
printf("%s", p[i]);
}
}
Later edit (as requested :-) )
int main( int, char**)
{
char *a[2]={"Do","While"};
myFunc( a, 2); // Could be myFunc( a, sizeof(a)/sizeof(char*));
// ...
return 0;
}
Too many stars - try
printf("%s",*P);
And you need %s format specifier - %c is just for single character.
If you want to print all strings, you need to pass number of strings in array and then print these strings from the loop.
Check the code suggested by Cătălin Pitiș. To pass the number of strings, you call function like this:
myFunc(a, sizeof(a)/sizeof(a[0]));
for( int i = 0; i < 2; i++ ) {
char* string = P[i];
printf( "%s", string );
}
And you shoud use some way of passing size of array into the function - either pass it as an int parameter,
int myFunc(char **P, int size)
{
for( int i = 0; i < size; i++ ) {
//whatever here
}
}
or always append a zero value to the array and only loop until you find that zero value.
char* array[] = { "String1", "String2", 0 };
Otherwise you will have hard to maintain code.
I like objective-c style nil (0) terminated arrays:
void myFunc(char **P)
{
while (*P) // loop to iterate over all strings until 0
printf("%s\n",*P++); // print and move to next element in array
}
int main()
{
char *a[]={"Do","While",0}; // added 0 to null terminate array,
myFunc(a); // kind of like string
}
Output:
Do
While
First, the good news: the type of a is equivalent to char **, so you are passing a valid parameter to myFunc().
The first problem is that %c is a format specifier that means print a single character. Since **P is an expression that evaluates to a single character, your first version does exactly what you told it to do. That isn't what you want.
The second version is close to syntactically correct. It should read printf("%s", *P), where *P is an expression that evaluates to a pointer to a nul-terminated ASCII string. In this case, it evaluates to "Do". This version won't print both strings.
Although it is true that the name of an array is the same as a pointer to its first element, that is a kind of "lie to students". Passing an array to a function does not and cannot convey the length of the array. To do that, you need either a second argument containing the length, or a convention like the nul-terminator on a string to indicate the end of the array. With that change, you can modify myFunc() to use a loop over the elements passed and print each one.
The problem in your code is that you want to print a string (char*) but you're giving it a char. Remember that P is an array of char*. When you de-reference it once, you get a char*; when you do it a second time, you just get the char at the beginning of the char*.
When you try to use the char value with the %s specifier, it treats the value as a pointer, and tries to dereference that value. Hence, it will try to print the "string" at the memory location X, where X is the value of the char (i.e. a value from 0 to 255). This gives you an access violation/segmentation fault (the error you see at runtime).
The best workarounds for this, as noted by Cătălin Pitiș and RBerteig, are to either:
pass another parameter to specify the length of the array
add an additional null at the end of the array.
if you don't want to keep and pass around array size::
int myFunc(char **);
main()
{
char *a[2]={"Do","While", NULL};
myFunc(a);
}
int myFunc(char **P)
{
if( !P )
return 0;
while(*P != NULL)
{
printf("%s",*P);
P++;
}
}
Wrong Answer: I think you may have to dereference P when you print it, although I could be wrong.
EDIT: I'm tired, it's 3 am here but I don't feel like sleeping so I'm here trying to answer questions. After reading the criticism, I reread the question and noticed that he does dereference P, but as is stated in another answer, it's dereferenced too much. When one wants to print a char string, one wants to pass a pointer as the char string is really an array.
Another EDIT: I would also like to point out that the person who asked the question made an edit after I answered and that when I first answered it didn't read "printf("%s",**P);" it read "printf("%s", P);" and the last part was bold.