same type, same address, but different value?

inmove · 10-09-2017, 04:33 AM

the code:

Code:

char *sendbuf = (char *)malloc(sizeof(char) * 1024);
char *sendbuftmp = sendbuf;
sendbuf = "hello world\0";                                                                                                                      

printf("sendbuftmp: %s, %d\n", sendbuftmp, strlen(sendbuftmp));
printf("sendbuf: %s, %d\n", sendbuf, strlen(sendbuf));

the output is:

sendbuftmp: , 0
sendbuf: hello world, 11

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sendbuf, sendbuftmp have the same address and the same type, but them have the different value, it confused me seriously.can anyone explain why, thx.

inmove · 10-09-2017, 04:44 AM

char *sendbuf = (char *)malloc(sizeof(char) * 1024);
sendbuf = "hello world\0";
char *sendbuftmp = sendbuf;

printf("sendbuftmp: %s, %d\n", sendbuftmp, strlen(sendbuftmp));
printf("sendbuf: %s, %d\n", sendbuf, strlen(sendbuf));

if i exchange the third and second line, the output is:

hello world,11
hello world,11

as expected.

pan64 · 10-09-2017, 06:34 AM

"hello world" is a string, it has its own address.

Code:

sendbuf = "hello world\0";

means sendbuf (which is a pointer) will get a new value, will point to the H of hello. The old value will be lost.

Code:

char *sendbuf = (char *)malloc(sizeof(char) * 1024);
printf("sendbuf: %s, %d\n", sendbuf, strlen(sendbuf));

is incorrect, because the 1024 chars were not initialized, so stren has no any meaning.