用malloc在其他函數中更改字符串

Question

當我使用malloc時，如何在其他函數中更改字符串？

in main: 

char *myString; 
changeString(myString); 


changeString(char *myString){ 

    myString = malloc((size) * sizeof(char)); 
    myString[1] = 'a'; 

} 

感謝您使用C

Answer 1

參數是按值傳遞。因此，要修改函數中的變量，您必須將指針傳遞給它。例如，int *至int和char **至char *。

void changeString(char **myString){ 
    // free(*myString); // add when myString is allocated using malloc() 
    *myString = malloc(2); 
    (*myString)[0] = 'a'; 
    (*myString)[1] = '\0'; 
} 

Answer 2

在main中分配內存，然後將一個指針傳遞給函數分配內存的開始。
同時將包含分配內存大小的變量傳遞給該函數，以便確保新文本不會溢出分配的內存。
修改的字符串可從main獲得。

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

void changeString(char *stringptr, int sz); 

int main(void) 
{ 
    const int a_sz = 16; 
    /* allocate memory for string & test successful allocation*/ 
    char *myString = malloc(a_sz); 
    if (myString == NULL) { 
     printf("Out of memory!\n"); 
     return(1); 
    } 

    /* put some initial characters into String */ 
    strcpy(myString, "Nonsense"); 
    /* print the original */ 
    printf("Old text: %s\n", myString); 

    /* call function that will change the string */ 
    changeString(myString, a_sz); 

    /* print out the modified string */ 
    printf("New text: %s\n", myString); 

    /* free the memory allocated */ 
    free(myString); 
} 

void changeString(char *stringptr, int sz) 
{ 
    /* text to copy into array */ 
    char *tocopy = "Sense"; 
    /* only copy text if it fits in allocated memory 
     including one byte for a null terminator */ 
    if (strlen(tocopy) + 1 <= sz) { 
     strcpy(stringptr, tocopy); 
    } 
}