C language string pointer to do function parameter example analysis
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See what's wrong with this code below?# include "stdio.h"
//#include "stdbool.h"
#include "string.h"
#include "stdlib.h"
#include "math.h"
void getMemory(char *p)
{
/*char *p = str*/
p = (char *)malloc(100);
strcpy(p,"hello world");
printf("p:%s\n",p);
}
int main()
{
printf("Enter main...\ n");
char *str = NULL;
printf("str:%p\n",str);
getMemory(str);
printf("%s\n",str);
if(str != NULL)
free(str);
return (0);
}
We look directly at the output, and the output looks like this.
Many people are not particularly clear about function parameters.
void getMemory(char *p)
{
/*char *p = str*/
p = (char *)malloc(100);
strcpy(p,"hello world");
printf("p:%s\n",p);
}
getMemory(str);
str is a pointer variable, that is, it stores a memory address, and this memory address points to the type char *"that is, string"
But when str is passed to getMemory (char * p), it passes a copy of str, not itself.
Since the copy is passed, the code operating in getMemory is also operating on this copy. When the function call ends, it is destroyed and recycled.
So str is still NULL.
How to modify?# include "stdio.h"
//#include "stdbool.h"
#include "string.h"
#include "stdlib.h"
#include "math.h"
void getMemory(char **p)
{
/*char **p = &str*/
*p = (char *)malloc(100);
strcpy(*p,"hello world");
printf("p:%s\n",*p);
}
int main()
{
printf("Enter main...\ n");
char *str = NULL;
printf("str:%p\n",str);
getMemory(&str);
printf("%s\n",str);
if(str != NULL)
free(str);
return (0);
}
Look at the output.
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