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I define a function 

int find(char *t, int len){
}

then i call it with 

value = "hello world";
rt = find(value, strlen(value));

it does not work, and show "error: too many arguments to function ‘find’"

the_drow
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why
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    The code posted will give an error about a missing semicolon. Try posting the EXACT code you're passing to the compiler -- there's probably some missing punctuation that is confusing the compiler and causing it to give a misleading error message. – Chris Dodd Jan 05 '11 at 05:51
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    Is this all done in one file or separate files? Has the definition or a prototype for `find)` been 'seen' before the call site (ie., do you do something like include a header that has `int find(char*,int);` in it)? – Michael Burr Jan 05 '11 at 05:51
  • thanks! Burr! I miss the header – why Jan 05 '11 at 06:02

4 Answers4

3
int find(char *t, int len){
}

might give a warning that function should return a value.

and if you add:

char* value = "hello world";
int rt = find(value, strlen(value));

It should work fine if the code is in a single file (as already pointed by Michael in comments) else you will have to specify the prototype of find function before calling it from a separate file.

Vikram.exe
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  1. There is a syntax error in your call, no ; after value = "hello world"

  2. Did you #include <string.h>?

milkypostman
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This error might occur when there is difference in arguments between function declaration and function definition.

iamnagaky
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There are two errors that I found in the above code .

You have to mention the 'return' keyword at the end of the function definition .

You have to declare the character pointer (char * value) while initialize the 'value' .

Usman
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