44 lines
1.3 KiB
Markdown
44 lines
1.3 KiB
Markdown
---
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title: Passing pointers to funtions
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---
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# Passing pointers to funtions
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C allows passing a pointer to a function. To achieve this, simply declare the parameters as pointer type.
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This way of passing references to functions is useful when you want to modify variables that are out of the scope of that function.
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```C
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// incorrect implementation of swap
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#include <stdio.h>
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void swap(int a, int b){
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int c;
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c = a;
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a = b;
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b = c;
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}
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int main(){
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int var1 = 10;
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int var2 = 20;
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swap(var1, var2);
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printf("Value of var1: %d \n", var1); // prints 10
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printf("Value of var2: %d \n", var2); // prints 20
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}
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```
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In this code example the swap function does not work as intended since it swaps two variables that exist only inside the scope of that function. To fix this we make a modification as shown below.
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```C
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// correct implementation of swap
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#include <stdio.h>
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void swap(int* a, int* b){
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int c = *a;
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*a = *b;
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*b = c;
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}
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int main(){
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int var1 = 10;
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int var2 = 20;
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swap(&var1, &var2);
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printf("Value of var1: %d \n", var1); // prints 20
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printf("Value of var2: %d \n", var2); // prints 10
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}
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```
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In the second code example you were able to change the values of the variables only because you were constantly de-referencing a pointer within the function instead of trying to change the values directly
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