为什么我的程序不打印出最终结果？
为什么当我运行此代码以输入用户的字符串输入时，为什么它不打印出最终结果？
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
/* Function Declerations */
/* Global Variables */
char *text = NULL;
int size;
int main(){
/* Initializing Global Variables */
printf("enter a number limit for text: ");
scanf("%d", &size);
/* Initial memory allocation */
text = (char *) malloc(size * sizeof(char));
if(text != NULL){
printf("Enter some text: \n");
scanf("%s", &text);
// scanf(" ");
// gets(text);
printf("You inputed: %s", text);
}
free(text);
return 0;
}
/* Function Details */
实际上，最终结果看起来像这样
enter a number limit for text: 20
Enter some text:
jason
why when i run this code to take in a string input by the user why does it not print out the final result ?
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
/* Function Declerations */
/* Global Variables */
char *text = NULL;
int size;
int main(){
/* Initializing Global Variables */
printf("enter a number limit for text: ");
scanf("%d", &size);
/* Initial memory allocation */
text = (char *) malloc(size * sizeof(char));
if(text != NULL){
printf("Enter some text: \n");
scanf("%s", &text);
// scanf(" ");
// gets(text);
printf("You inputed: %s", text);
}
free(text);
return 0;
}
/* Function Details */
in fact the end result looks like this
enter a number limit for text: 20
Enter some text:
jason
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评论（2）
text已经是char *，因此您不必将＆amp; text转换为scanf，而是仅文本。scanf将指针作为参数以修改指向值，但是如果将char ** **作为参数，您将修改到字符串的指针而不是尖的字符串textis already achar *, so you don't have to pass&texttoscanf, but onlytext.scanftakes a pointer as argument in order to modify the pointed value, but if you pass achar **as an argument, you will modify the pointer to the string instead of the pointed string您只需要从＆amp; text 中删除地址，因为文本是指针，字符串总是指向第一个字符。
you just have to remove the address from &text because text is a pointer and the string always pointe to the first character.