[Warning]assignment makes integer from pointer without a cast

[Shane Fitz]

The title is the warning I get when I try to run the following function

char * setup_mutations()
{
	char *result = malloc(sizeof(char));
	int i;
	char c, current, string[MAX_STRING];
	FILE * fp;
	
	if((fp = fopen("mutations.txt","r"))==NULL){
		printf("Mutations file not available\n");
		exit(2);
	}
	
	for(i=0; i<256; i++){
		//result[i] = malloc(sizeof(char));
		result[i] = '\0';
	}
	
	while ( (c = fgetc(fp) ) != EOF ){
		current = c;
		
		for(c=fgetc(fp); c!=' '; c=fgetc(fp)){
			
		}
		
		i=0;
		
		for(c=c; c!='\n'; c=fgetc(fp)){
			string[i] = c;
			i++;
		}
		i = current;
		result[i] = string;
	}
	
	if((fclose(fp))==-1)printf("Mutations file didn't close properly\n");
	return result;
};

It comes from the line 32. Any help that helps me fix this problem and understand why it has arisen is appreciated.

Thanks

[donbock]

1. Line 3 establishes result as a pointer to a single char; however line 15 seems to think that result points to 256 chars. It doesn't.
2. Lines 21 and 27. Consider what happens if fgetc returns EOF.
3. Line 21. It is customary for no-body for loop to have a single bare semicolon as the body.
4. Line 28. How do you know you won't write past the end of string?
5. Line 32. result is a pointer to a single char; thus result[i] is a char. This line tries to store a pointer-to-char there. That's why you get that warning.