This what wrong ah, have without bosses to solve!!!!!!

CodePudding user response:

Swap not declare, your function?

CodePudding user response:

Put your swap function int main () to the front,

Or, as the upstairs said, in the int main () with a disclaimer:

Void swap (int * p_1, p_2 int *)

Also, when you swap function has no return value, best and void,

CodePudding user response:

Upstairs said is truth, if you don't forward declaration function, is going to put it in front of the main ()

CodePudding user response:

Line 14 to 20 rows to the back of the line 1

CodePudding user response:

 # include & lt; Stdio. H> 
Void swap (int * p_1, p_2 int *);//-- -- -- -- -- - add swap () function statement -- -- -- -- -- -- -- -- -- 
Int main () 
{
Int a, b; 
P1, int * * p2. 
The scanf (" % d % d ", & amp; A, & amp; B);//-- -- -- -- -- - don't have a comma between two % d -- -- -- -- -- -- -- -- -- 
P1=& amp; a; 
The p2=& amp; b; 
If (a & gt; B) 
{
Swap (p1, p2);//-- -- -- -- -- - pay attention to the function of the parameter type is a pointer, * (p1 and p2 to p1 and p2 - -- -- -- -- -- -- -- -- -- -- 
} 
Printf (" % d, % d \ n ", * (p1, * (p2); 
Printf (" % d, % d \ n ", a, b); 
return 0; 
} 

Void swap (int * p_1, p_2 int *) 
{
Int p; 
P=* p_1; 
* p_1=* p_2; 
* p_2=p; 
} 

C + + environment debugging VS2015