我对使用感到困惑%c
and %s
在以下 C 程序中:
#include <stdio.h>
void main()
{
char name[] = "siva";
printf("%s\n", name);
printf("%c\n", *name);
}
Output:
siva
s
为什么显示字符%c需要用指针,而字符串则不需要指针
我运行时遇到错误
printf("%c\n", name);
我收到这个错误:
str.c: In function ‘main’:
str.c:9:2: warning: format ‘%c’ expects type ‘int’, but argument 2 has type ‘char *’
如果你尝试这样做:
#include<stdio.h>
void main()
{
char name[]="siva";
printf("name = %p\n", name);
printf("&name[0] = %p\n", &name[0]);
printf("name printed as %%s is %s\n",name);
printf("*name = %c\n",*name);
printf("name[0] = %c\n", name[0]);
}
输出是:
name = 0xbff5391b
&name[0] = 0xbff5391b
name printed as %s is siva
*name = s
name[0] = s
所以“name”实际上是一个指向内存中字符数组的指针。如果您尝试读取 0xbff5391b 处的前四个字节,您将看到“s”、“i”、“v”和“a”
Location Data
========= ======
0xbff5391b 0x73 's' ---> name[0]
0xbff5391c 0x69 'i' ---> name[1]
0xbff5391d 0x76 'v' ---> name[2]
0xbff5391e 0x61 'a' ---> name[3]
0xbff5391f 0x00 '\0' ---> This is the NULL termination of the string
要打印字符,您需要将字符的值传递给printf
。该值可以参考为name[0]
or *name
(因为对于数组name = &name[0]
).
要打印字符串,您需要将指向该字符串的指针传递给printf
(在这种情况下name
or &name[0]
).
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