Thread: "cannot convert parameter 1 from 'const char * []' to 'char *'" error

Hi,

I've some problems to get over, first of all, i keep getting this error and couldn't get it solved. here is the code and the error i get:

the error: error C2664: 'strtok' : cannot convert parameter 1 from 'const char []' to 'char *'

Code:

#include <stdio.h>
#include <string.h>
#include <ctype.h>
void countletters(const char a[]);

int main() {
	char string[80];
	puts("Text Analysis 1.0");
	puts("Please enter a string to be analyzed");
	scanf("%s",string);
	countletters(string);
	return 0;
}

void countletters(const char a[]) {
	int letters[26]={0};
	int length;
	int wordlength[11];
	char *token;
	int i,x;
	length = (int)strlen(a);
	printf("Lenght of the text is: %d\n",length);
	puts("Number of occurences of each letter:");
	for (i=0; i<length; ++i) {
		x=tolower(a[i]);
		x -= 97;
		++letters[x];
	}
	for (i=0;i<26;++i) {
		printf("%c:%d\t",97+i,letters[i]);
		if (!((i+2)%9)) printf("\n");
	}
	puts("The lengths of the words in the sentence:");
	token = strtok(a," ");
	while (token != NULL) {
		x=strlen(token);
		++wordlength[x];
		strtok(NULL, " ");
	}
	for (i=1;i<11;++i)
		printf("%d%s%d\n",i," letter words: ",wordlength[i]);
}

Second question: This piece of code is from a high school programming contest, the question is to determine the output by mind. I answered the question wrong, give it a try and please explain the output comes out you chose(by the way, don't mind the preprocessor statements, return 0, etc):

Code:

void f(int p[],int q[])
{
	int a[5]={1,2,3,4,5};
	static int b[5]={-1,-2,-3,-4,-5};
	p = a;
	q = b;
}
main()
{
	int i,j, p[5]={2,4,6,8,10}, q[5]={1,3,5,7,9};
	f(p,q);
	for(i=0;i<5;i++) p[i] += q[i];
	for(i=0;i<5;i++) printf("%d ",p[i]);
}

a) 1 2 3 4 5
b) 2 5 8 11 14
c) 0 1 2 3 4
d) 3 7 11 15 19
e) none of the above

Thank you for your help and kind concern!

For the second question it looks like d is the winner to me. The f() function has no effect on the main() function. Not sure why j is there either. Of course, d isn't quite right either because the printf() doesn't have a space in the format string so the numbers would come out in one long string: 37111519

If there truly isn't a space in the printf() format string I would have to go with e instead.

what i couldn't get is that, you say the function has no effect on main, but


Code:
p = a;
q = b;

shouldn't these change the values of the first elements of the arrays?

Nope. f() is only setting its local variables to those arrays. It's the same as:

Code:

void f(int a)
{
  a = 5;
}
int main(void)
{
  int a = 3;
  f(a);
  printf("%d\n", a);
  return 0;
}

That program will still print 3. If you wanted to change the value of a in main() you'd have to pass a pointer to a instead of the value of a. The same goes with your mind question. You'd have to pass a pointer to the pointer if you want to change the pointer in main().

Originally Posted by kolistivra

AFAIK, arrays are passed by reference, unlike other data types.

Kind of. Copies of pointers to the first element in the array are passed. Your code just excanges this pointers. the values are not touched.
Kurt
EDIT:
if you want to see an output of all zeros, you have to code f() like this

Code:

void f(int p[],int q[]) {
        int i;
	int a[5]={1,2,3,4,5};
	static int b[5]={-1,-2,-3,-4,-5};
        
        for ( i = 0; i < 5; ++i )
	    p[i] = a[i];
        for ( i = 0; i < 5; ++i )
	    q[i] = b[i];
}

AFAIK, arrays are passed by reference, unlike other data types.

An array decays to a pointer to its first element when it is passed as an argument, and this pointer is passed by value.