Question

This is a newbie question. To create a formatted C string, I use printf, like:

int n = 10;
printf("My number is %i", 10);

但是,如何:

int n = 10
char *msg = "My number is %i", 10;
printf(msg);

我怎么能够把由此产生的格式拼凑成一个变数? 我想“这个数字是10”。

Answer 1

http://linux.die.net/man/3/snprintf"rel=“noretinger”>snprintf(:

int n = 10;
char bla[32];   // Use an array which is large enough 
snprintf(bla, sizeof(bla), "My number is %i", n);

Do not use sprintf(); 这与snprintf相似,但并未进行任何缓冲尺寸的检查,因此这被视为安全漏洞——当然,你可能总是分配足够的记忆,但你可能忘记,有时会给它留下巨大的安全漏洞。



如果你想把记忆分配给你,你可使用http://linux.die.net/man/3/asprintf” rel=“noreferer”asprintf(。 相反:

int n = 10;
char *bla;
asprintf(&bla, "My number is %i", n);
// do something with bla
free(bla); // release the memory allocated by asprintf.

Answer 2

参看sprintf(。

int ret;
int n=10;
char msg[50];  /* allocate some space for string */

/* Creates string like printf, but stores in msg */
ret = sprintf(msg,"My number is %i",n); 
printf(msg);

Answer 3

使用<条码>。

int n=10
char *msg ="My number is %i";
char bla[32];   // Use an array which is large enough 
sprintf(bla, msg, n);

Answer 4

You would need to use something like sprintf http://www.rohitab.com/discuss/topic/11505-sprintf-tutorial-in-c/

it s used basically like this (remember to malloc the msg variable first)

char* msg;
int ret = sprintf(msg,"My number is %i",10);
printf(msg);

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