Question

I want to change the contents of a constant-character-array(const array[64]).
Below is my code.
My Question is, why does the constant character array doesn t change(not reflected back), when passed to the function as constant character pointer(const char *append)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int function(char *d,const char *append)
{
 append = d; //changing the location of append.
 printf ("%s
",append); //displays as sachintendulkar.
}

int main()
{
    char *d = NULL;
    const char append[]={ s , a , c , h , i , n };
    d = calloc(sizeof(char),sizeof(append));
    strcpy(d,append);
    strcat(d,"tendulkar"); //appending
    function(d,append);
    printf ("%s
",append); //Its displays as sachin instead of sachintendulkar???
}

Answer 1

函数参数被值传递,当您在 函数() 函数 > 内为指针指定新值 append 后,在函数外没有出现任何现象。

并不清楚你试图做什么... 数据不变的点当然是,你不应该改变它。

Answer 2

参数名称与主变量相同只是巧合。名称之间没有关联。

您的函数工作与您的工作功能相同

int function(char *x, const char *y)
{
 y = x; //changing the location of y.
 printf ("%s
", y); //displays as sachintendulkar.
}

您不会期望该函数会改变主体内部的值。

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