赋值使指针从整数变成整数，而不进行强制转换[-Wint-conversion]

Question

i am new to c, i am trying to make a simple code in which i replace the vowels of an input with the * character... i am having trouble with the array (i am mainly familiar with python only using c now.)

#include <stdio.h>

 char* main() {
 int c;
 char* vowels[] = {'a', 'e', 'i', 'o', 'u'};

 while( 1 ) {   // means: true
  c = getchar();
  if(c == EOF) break;
  if(c = vowels) putchar('*'); 
  else putchar(c);
}
}

i keep receiving the same error message: novowels.c:5:41: note: (near initialization for 'vowels[4]') novowels.c:10:12: warning: assignment makes integer from pointer without a cast [-Wint-conversion] if(c = vowels) putchar('*'); ^

.... please help, i can't quite figure out how do complete this program of replacing vowels with *

Answer 1

Firstly, char *main() is illegal in C. It should be int main() .

Secondly, 'a' and the like are integer constants in C. You cannot use them to initialize and array of char * elements. If you wanted a char array you should have declared it as

const char vowels[] = {'a', 'e', 'i', 'o', 'u'};

If you wanted a string array you should have declared it as

const char *const vowels[] = {"a", "e", "i", "o", "u"};

Finally, it is completely unclear what you were trying to say by c = vowels . If you want to check whether c is in vowels array, keep in mind that there's no core language feature that can do it for you (and certainly not the = operator). You have to implement the check manually or use an applicable library feature.

For example, one possible way to check it is

const char vowels[] = {'a', 'e', 'i', 'o', 'u'};
...
if (memchr(vowels, c, sizeof vowels) != NULL)
  putchar('*');

Alternatively, it can be expressed as

const char *const vowels = "aeiou";
...
if (strchr(vowels, c) != NULL)
  putchar('*');

Answer 2

First of all, your program contains a typo. == is comparison whereas = is assignment. Instead of this line

if(c = vowels) putchar('*');

you probably meant to write

if(c == vowels) putchar('*');

but even that is incorrect. You cannot use the == operator to test for membership. To test for membership, you could use the strchr() function from <string.h> :

if (memchr(vowels, c, sizeof vowels) != NULL) putchar('*');

You can read here about what memchr does.

Lastly, as AnT already remarked, main shouldn't return a char* . The main function must return an int in a strictly conforming C program. Change the line

char* main() {

to

int main() {

to fix this.

Answer 3

The problem is that you are not comparing anything in c = vowels . You are assigning to c the address of the array vowels (which is a pointer indeed).

Probably you wanted to check equality between c and one of the vowels but you must use the [] operator since vowels is an array, an index, and a different operator (comparison, which is == ). Something like:

if (c == vowels[2])

but then vowels should be a char[] not a char*[] .